Lecture Notes on Integral Calculus - Ugrad.math.ubc.ca
Lecture Notes on Integral Calculus - Ugrad.math.ubc.ca
Lecture Notes on Integral Calculus - Ugrad.math.ubc.ca
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2.3 The sigma notati<strong>on</strong>In order to short-hand the <strong>math</strong>emati<strong>ca</strong>l exressi<strong>on</strong> of the sum of a regular sequence, a c<strong>on</strong>venientnotati<strong>on</strong> is introduced.Definiti<strong>on</strong> (Σ sum): The sum of the first k terms of a sequence generated by the sequencegenerator f(i) <strong>ca</strong>n be denoted byS k = f(1) + f(2) + · · · + f(k) ≡k∑f(i)i=1where the symbol Σ (the Greek equivalent of S reads “sigma”) means “take the sum of”,the general expressi<strong>on</strong> for the terms to be added or the sequence generator f(i) is <strong>ca</strong>lled thesummand, i is <strong>ca</strong>lled the summati<strong>on</strong> index, 1 and k are, respectively, the starting and theending indices of the sum.Thus,k∑f(i)i=1means <strong>ca</strong>lculate the sum of all the terms generated by the sequence generator f(i) for all integersstarting from i = 1 and ending at i = k.Note that the value of the sum is independent of the summati<strong>on</strong> index i, hence i is <strong>ca</strong>lled a“dummy” variable serving for the sole purpose of running the summati<strong>on</strong> from the startingindex to the ending index. Therefore, the sum <strong>on</strong>ly depends <strong>on</strong> the summand and both thestarting and the ending indices.∑Example 5: Express the sum S k = k i 2 in an expanded form.i=3Soluti<strong>on</strong> 5: The sequence generator is f(i) = i 2 . Note that the starting index is not 1 but 3!.Thus, the 1st term is f(3) = 3 2 . The subsequent terms <strong>ca</strong>n be determined accordngly. Thus,S k =k∑i 2 = f(3) + f(4) + f(5) + · · · + f(k) = 3 2 + 4 2 + 5 2 + · · · + k 2 .i=3An easy check for a mistake that often occurs. If you still find the “dummy” variable i in anexpanded form or in the final evaluati<strong>on</strong> of the sum, your answer must be WRONG.2.4 Gauss’s formula and other formulas for simple sumsLet us return to Examples 1 and 2 about the total number of logs in a triangular pile. Let’sstart with a pile of 4 layers. Imaging that you could (in a “thought-experiement”) put an4
identi<strong>ca</strong>l pile with up side down adjacent to the original pile, you obtain a pile that c<strong>on</strong>tainstwice the number of logs that you want to <strong>ca</strong>lculate (see figure).The advantage of doing this is that, in this double-sized pile, each layer c<strong>on</strong>tains an equalnumber of logs. This number is equal to number <strong>on</strong> the 1st (top) layer plus the number <strong>on</strong> the4th (bottom) layer. In the mean time, the height of the pile remains unchanged (4). Thus, thenumber in this double-sized pile is 4 × (4 + 1) = 20. The sum S 4 is just half of this numberwhich is 10.Let’s apply this idea to finding the formula in the <strong>ca</strong>se of k layers. Note that(Original) S k = 1 + 2 + · · · + k − 1 + k. (2)(Inversed) S k = k + k − 1 + · · · + 2 + 1. (3)(Adding the two) 2S k = (k + 1) + (k + 1) + · · · + (k + 1) + (k + 1) = k(k + 1).} {{ }(4)k terms in totalDividing both sides by 2, we obtain Gauss’s formula for the sum of the first k positive integers.S k =k∑i=1i = 1 + 2 + · · · + k = 1 k(k + 1). (5)2This actaully answered the problem in Example 2. The answer to Example 1b is even simpler.S 50 =50∑i=1i = 1 × 50 × 51 = 1275.2The following are two important simple sums that we shall use later. One is the sum of thefirst k integers squared.S k =k∑i=1i 2 = 1 2 + 2 2 + · · · + k 2 = 1 k(k + 1)(2k + 1). (6)6The other is the sum of the first k integers cubed.S k =k∑[ ] 2 1i 3 = 1 3 + 2 3 + · · · + k 3 =2 k(k + 1) . (7)i=1We shall not illustrate how to derive these formulas. You <strong>ca</strong>n find it in numerous <strong>ca</strong>lculus textbooks.To prove that these formulas work for arbitrarily large integers k, we <strong>ca</strong>n use a method <strong>ca</strong>lled<strong>math</strong>emati<strong>ca</strong>l inducti<strong>on</strong>. To save time, we’ll just outline the basic ideas here.5
The <strong>on</strong>ly way we <strong>ca</strong>n prove things c<strong>on</strong>cerning arbitrarily large numbers is to guarantee thatthis formula must be correct for k = N + 1 if it is correct for k = N. This is like tryingto arrange a string/sequence of standing dominos. To guarantee that all the dominos in thestring/sequence fall <strong>on</strong>e after the other, we need to guarantee that the falling of each dominowill necessarily <strong>ca</strong>use the falling of the subsequent <strong>on</strong>e. This is the essence of proving that ifthe formula is right for k = N, it must be right for k = N + 1. The last thing you need to dois to knock down the first <strong>on</strong>e and keep your fingers crossed. Knocking down the first dominois equivalent to proving that the formula is correct for k = 1 which is very easy to check in all<strong>ca</strong>ses (see Keshet’s notes for a prove of the sum of the first k integers squared).2.5 Important rules of sigma sumsRule 1: Summati<strong>on</strong> involving c<strong>on</strong>stant summand.k∑i=1c = c}+ c +{{· · · +}c = kc.k terms in totalNote that: the total # of terms=ending index - starting index +1.Rule 2: C<strong>on</strong>stant multipli<strong>ca</strong>ti<strong>on</strong>: multiplying a sum by a c<strong>on</strong>stant is equal to multiplying eachterm of the sum by the same c<strong>on</strong>stant.k∑c f(i) =i=1k∑cf(i)i=1Rule 3: Adding two sums with identi<strong>ca</strong>l starting and ending indices is equal to the sum of sumsof the corresp<strong>on</strong>ding terms.k∑f(i) +i=1k∑g(i) =i=1k∑[f(i) + g(i)].i=1Rule 4: Break <strong>on</strong>e sum into more than <strong>on</strong>e pieces.k∑f(i) =i=1n∑k∑f(i) + f(i),i=1 i=n+1(1 ≤ n < k).Example 7: Let’s go back to solve the sum of the first k odd numbers.Soluti<strong>on</strong> 7:k∑k∑ k∑O k = 1 + 3 + · · · + (2k − 1) = (2i − 1) = (2i) − 1 = 2111k∑i − k,16
where Rules 1, 2, and 3 are used in the last two steps. Using the known formulas, we obtainO k =k∑(2i − 1) = k(k + 1) − k = k 2 .1Example 8: Let us now go back to solve Example 4. It is the sum of the first k products ofpairs of subsequent integers.Soluti<strong>on</strong> 8:P k = 1 · 2 + 2 · 3 + · · · + k · (k + 1) =k∑i(i + 1) =1k∑[i 2 + i] =where Rule 3 was used in the last step. Applying the formulas, we learnedP k =k∑11k∑i 2 +i(i + 1) = 1 6 k(k + 1)(2k + 1) + 1 2 k(k + 1) = 1 k(k + 1)(k + 2).3This is another simple sum that we <strong>ca</strong>n easily remember.∑Example 9: Calculate the sum S = k (i + 2) 3 .11k∑i,Soluti<strong>on</strong> 9: This problem <strong>ca</strong>n be solved in two different ways. The first is to expand thesummand f(i) = (i + 2) 3 which yieldS =k∑(i + 2) 3 =1k∑(i 3 + 6i 2 + 12i + 8) =1k∑i 3 + 61k∑i 2 + 12We <strong>ca</strong>n solve the resulting three sums separately using the known formulas.11k∑i + 8k.But there is a better way to solve this. This involves substituting the summati<strong>on</strong> index. Wefind it easier to see how substituti<strong>on</strong> works by expanding the sum.S =k∑(i + 2) 3 = 3 3 + 4 3 + · · · + k 3 + (k + 1) 3 + (k + 2) 3 .i=1We see that this is simply a sum of integers cubed. But the sum does not start at 1 3 and endat k 3 like in the formula for the sum of the first k integers cubed.Thus we re-write the sum with a sigma notati<strong>on</strong> with an new index <strong>ca</strong>lled l which starts atl = 3 and ends at l = k + 2 (there is no need to change the symbol for the index, you <strong>ca</strong>n keep<strong>ca</strong>lling it i if you do not feel any c<strong>on</strong>fusi<strong>on</strong>). Thus,1S =k∑∑k+2(i + 2) 3 = 3 3 + 4 3 + · · · + k 3 + (k + 1) 3 + (k + 2) 3 = l 3 .i=1l=37
We just finished doing a substituti<strong>on</strong> of the summati<strong>on</strong> index. It is equivalent to replacing iby l = i + 2. This relati<strong>on</strong> also implies that i = 1 ⇒ l = 3 and i = k ⇒ l = k + 2. This isactually how you <strong>ca</strong>n determine the starting and the ending values of the new index.Now we <strong>ca</strong>n solve this sum using Rule 4 and the known formula.S =k∑ ∑k+2∑k+22∑[ ] 2 [ ] 2 1 1(i+2) 3 = l 3 = l 3 − l 3 =2 (k + 2)(k + 3) −1 2 −2 3 =2 (k + 2)(k + 3) −9.i=1l=3 l=1 l=12.6 Appli<strong>ca</strong>ti<strong>on</strong>s of sigma sumThe area under a curveWe know that the area of a rectangle with length l and width w is A rect = w · l.Starting from this formula we <strong>ca</strong>n <strong>ca</strong>lculate the area of a triangle and a trapezoid. This isbe<strong>ca</strong>use a triangle and a trapezoid <strong>ca</strong>n be transformed into a rectangle (see Figure). Thus, fora triangle of height h and base length bA trig = 1 2 hb.Similarly, for a trapezoid with base length b, top length t, and height hA trap = 1 h(t + b).2Following a very similar idea, the sum of a trapezoid-shaped pile of logs with t logs <strong>on</strong> toplayer, b logs <strong>on</strong> the bottom layer, and a height of h = b − t + 1 layers (see figure) isb∑i=ti = t + (t + 1) + · · · + (b − 1) + b = 1 2 h(t + b) = 1 (b − t + 1)(t + b). (8)2Now returning to the problem of <strong>ca</strong>lculating the area. Another important formula is for thearea of a circle of radius r.A circ = πr 2 .Now, <strong>on</strong>ce we learned sigma and/or integrati<strong>on</strong>, we <strong>ca</strong>n <strong>ca</strong>lculate the area under the curve ofany functi<strong>on</strong> that is integrable.Example 10: Calculate the area under the curve y = x 2 between 0 and 2 (see figure).Soluti<strong>on</strong> 10: Remember always try to reduce a problem that you do not know how to solveinto a problem that you know how to solve.8
Let the area be A. Let’s divide A into 3 rectangles of equal width w = 2/3. Thus,{w[xA ≈ w·h 1 +w·h 2 +w·h 3 =2 0 + x2 1 + x2 2 ] = 1 3 [02 + ( )1 2+ ( 2 23 3)], left − end approx.;w[x 2 1 + x2 2 + x2 3 ] = 1[( )1 2+ ( )2 2+ ( 3 23 3 3 3)], right − end approx..By using these rectangles, we introduced large errors in our estimates using the heights based<strong>on</strong> both the left and the right endpoints of the subintervals. To increase accurary, we need toincrease the number of rectangles by making each <strong>on</strong>e thinner. Let us now divide A into nrectangles of equal width w = 2/n. Thus, using the height based <strong>on</strong> the right endpoint of eachsubinterval, we obtainA ≈ S n = A 1 + A 2 + · · · + A n =n∑w · h i = wi=1n∑f(x i ) = 2 ni=1n∑x 2 i .i=1It is very important to keep a clear account of the height of each rectangle. In this <strong>ca</strong>se,h i = f(x i ) = x 2 i . Thus, the key is finding the x-coordinate of the right-end of each rectangle.For rectangles of of equal width, x i = x 0 + iw = x 0 + i∆x/n, where ∆x is the length of theinterval x 0 is the left endpoint of the interval. x 0 = 0 and ∆x = 2 for this example. Therefore,x i = i(2/n). Substitute into the above equati<strong>on</strong>A ≈ S n = 2 nn∑x 2 i . = 2 ni=1n∑i=1( ) 2 2i= 8 n∑i 2 = 8 1n n 3 n 3 6 n(n + 1)(2n + 1) = 4 (n + 1)(2n + 1).3 n 2 i=1Note that if we divide the area into infinitely many retangles with a width that is infinitelysmall, the approximate becomes accurate.A = limn→∞S n = lim3n→∞4The volume of a solid revoluti<strong>on</strong> of a curve(n + 1)(2n + 1)n 2 = 8 3 .We know the volume of a rectangular block of height h, width w, and length l isV rect = w · l · h.The volume of a cylinder of thichness h and radius rV cyl = A cross · h = πr 2 h.Example 11: Calculate the volume of the bowl-shaped solid obtained by rotating the curvey = x 2 <strong>on</strong> [0, 2] about the y-axis.9
2.7 The sum of a geometric sequence3 The Definite <strong>Integral</strong>s and the Fundamental Theorem3.1 Riemann sumsDefiniti<strong>on</strong> 1: Suppose f(x) is finite-valued and piecewise c<strong>on</strong>tinuous <strong>on</strong> [a, b]. Let P ={x 0 = a, x 1 , x 2 , . . . , x n = b} be a partiti<strong>on</strong> of [a, b] into n subintervals I i = [x i−1 , x i ] of width∆x i = x i − x i−1 , i = 1, 2, . . . , n. (Note in a special <strong>ca</strong>se, we <strong>ca</strong>n partiti<strong>on</strong> it into subintervalsof equal width: ∆x i = x i − x i−1 = w = (b − a)/n for all i). Let x ∗ i be a point in I i such thatx i−1 ≤ x ∗ i ≤ x i . (Here are some special ways to choose x ∗ i : (i) left endpoint rule x∗ i = x i−1 ,and (ii) the right endpoint rule x ∗ i = x i ).The Riemann sums of f(x) <strong>on</strong> the interval [a, b] are defined by:n∑n∑R n = h i ∆x i = f(x ∗ i )∆x ii=1which approximate the area between f(x) and the x-axis by the sum of the areas of n thinrectangles (see figure).Example 1: Approximate the area under the curve of y = e x <strong>on</strong> [0, 1] by 10 rectangles ofequal width using the left endpoint rule.Soluti<strong>on</strong> 1: We are actually <strong>ca</strong>lculating the Riemann sum R 10 . The width of each rectangleis ∆x i = w = 1/10 = 0.1 (for all i). The left endpoint of each subinterval is x ∗ i = x i − 1 =(i − 1)/10. Thus,R 10 =10∑i=1f(x ∗ i )∆x i =10∑i=1i=1e (i−1)/10 (0.1) = 0.19∑e j/10 = 0.11 − e ≈ 1.6338.1 − e1/10 j=03.2 The definite integralDefiniti<strong>on</strong> 1: Suppose f(x) is finite-valued and piecewise c<strong>on</strong>tinuous <strong>on</strong> [a, b]. Let P = {x 0 =a, x 1 , x 2 , . . . , x n = b} be a partiti<strong>on</strong> of [a, b] with a length defined by |p| = Max 1≤i≤n {∆x i }(i.e. the l<strong>on</strong>gest of all subintervals). The definite integral of f(x) <strong>on</strong> [a, b] is∫ baf(x)dx =lim R n =|p|→0; n→∞lim|p|→0; n→∞n∑f(x ∗ i )∆x iwhere the symbol ∫ means “to integrate”, the functi<strong>on</strong> f(x) to be integrated is <strong>ca</strong>lled theintegrand, x is <strong>ca</strong>lled the integrati<strong>on</strong> variable which is a “dummy” variable, a and b are,i=110
espectively, the lower (or the starting) limit and the upper (or the ending) limit of the integral.Thus,∫ baf(x)dxmeans integrate the functi<strong>on</strong> f(x) starting from x = a and ending at x = b.Example 2: Calculate the definite integral of f(x) = x 2 <strong>on</strong> [0.2]. (This is Example 10 of<str<strong>on</strong>g>Lecture</str<strong>on</strong>g> 1 reformulated in the form of a definite integral).Soluti<strong>on</strong> 2:I =∫ 20x 2 dx = limn→∞n∑( 2ini=1) 2} {{ }f(x ∗ i )( ) 2n} {{ }∆x i( ) n∑ 8= lim i 2 = limn→∞ n 3 i=13n→∞4(n + 1)(2n + 1)n 2 = 8 3 .Important remarks <strong>on</strong> the relati<strong>on</strong> between an area and a definite integral:• An area, defined as the physi<strong>ca</strong>l measure of the size of a 2D domain, is always n<strong>on</strong>negative.• The value of a definite integral, sometimes also referred to as an “area”, <strong>ca</strong>n be bothpositive and negative.• This is be<strong>ca</strong>use: a definite integral = the limit of Riemann sums. But Riemann sumsare defined asn∑n∑R n = (area of i th rectangle) = f(x ∗ i ) ∆x} {{ } i . }{{}i=1i=1height width• Note that both the hieght f(x ∗ i ) and the width ∆x i <strong>ca</strong>n be negative implying that R n<strong>ca</strong>n have either signs.3.3 The fundamental theorem of <strong>ca</strong>lculus3.4 Areas between two curves11
4 Appli<strong>ca</strong>ti<strong>on</strong>s of Definite <strong>Integral</strong>s: I4.1 Displacement, velocity, and accelerati<strong>on</strong>4.2 Density and total mass4.3 Rates of change and total change4.4 The average value of a functi<strong>on</strong>12
5 DifferentialsDefiniti<strong>on</strong>: The differential, dF , of any differentiable functi<strong>on</strong> F is an infinitely small incrementor change in the value of F .Remark: dF is measured in the same units as F itself.Example: If x is the positi<strong>on</strong> of a moving body measured in units of m (meters), then itsdifferential, dx, is also in units of m. dx is an infinitely small increment/change in the positi<strong>on</strong>x.Example: If t is time measured in units of s (sec<strong>on</strong>ds), then its differential, dt, is also in unitsof s. dt is an infinitely small increment/change in time t.Example: If A is area measured in units of m 2 (square meters), then its differential, dA, alsois in units of m 2 . dA is an infinitely small increment/change in area A.Example: If V is volume measured in units of m 3 (cubic meters), then its differential, dV ,also is in units of m 3 . dV is an infinitely small increment/change in volume V .Example: If v is velocity measured in units of m/s (meters per sec<strong>on</strong>d), then its differential,dv, also is in units of m/s. dv is an infinitely small increment/change in velocity v.Example: If C is the c<strong>on</strong>centrati<strong>on</strong> of a biomolecule in our body fluid measured in units ofM (1 M = 1 molar = 1 mole/litre, where 1 mole is about 6.023 × 10 23 molecules), thenits differential, dC, also is in units of M. dC is an infinitely small increment/change in thec<strong>on</strong>centrati<strong>on</strong> C.Example: If m is the mass of a rocket measured in units of kg (kilograms), then its differential,dm, also is in units of kg. dm is an infinitely small increment/change in the mass m.Example: If F (h) is the culmulative probability of finding a man in Canada whose height issmaller than h (meters), then dF is an infinitely small increment in the probability.Definiti<strong>on</strong>: The derivative of a functi<strong>on</strong> F with respect to another functi<strong>on</strong> x is defined asthe quotient between their differentials:13
dFdx=an infinitely small rise in Fan infinitely small run in x .Example: Velocity as the rate of change in positi<strong>on</strong> x with respect to time t <strong>ca</strong>n be expressedasv =infinitely small change in positi<strong>on</strong>infinitely small time interval= dxdt .Remark: Many physi<strong>ca</strong>l laws are correct <strong>on</strong>ly when expressed in terms of differentials.Example: The formula (distance) = (velocity) × (time interval) is true either when thevelocity is a c<strong>on</strong>stant or in terms of differentials, i.e., (an infinitely small distance) =(velocity) × (an infinitely short time interval). This is be<strong>ca</strong>use in an infinitely short timeinterval, the velocity <strong>ca</strong>n be c<strong>on</strong>sidered a c<strong>on</strong>stant. Thus,which is nothing new but v(t) = dx/dt.dx = dx dt = v(t)dt,dtExample: The formula (work) = (force) × (distance) is true either when the forceis a c<strong>on</strong>stant or in terms of differentials, i.e., (an infinitely small work) = (force) ×(an infinitely small distance). This is be<strong>ca</strong>use in an infinitely small distance, the force <strong>ca</strong>nbe c<strong>on</strong>sidered a c<strong>on</strong>stant. Thus,dW = fdx = f dx dt = f(t)v(t)dt,dtwhich simply implies: (1) f = dW/dx, i.e., force f is nothing but the rate of change in workW with respect to distance x; (2) dW/dt = f(t)v(t), i.e., the rate of change in work W withrespect to time t is equal to the product between f(t) and v(t).Example: The formula (mass) = (density) × (volume) is true either when the densityis c<strong>on</strong>stant or in terms of differentials, i.e., (an infinitely small mass) = (density) ×14
(volume of an infinitely small volume). This is be<strong>ca</strong>use in an infinitely small piece ofvolume, the density <strong>ca</strong>n be c<strong>on</strong>sidered a c<strong>on</strong>stant. Thus,dm = ρdV,which implies that ρ = dm/dV , i.e., density is nothing but the rate of change in mass withrespect to volume.6 The Chain Rule in Terms of DifferentialsWhen we differentiate a composite functi<strong>on</strong>, we need to use the Chain Rule. For example,f(x) = e x2 is a composite functi<strong>on</strong>. This is be<strong>ca</strong>use f is not an exp<strong>on</strong>ential functi<strong>on</strong> of x butit is an exp<strong>on</strong>ential functi<strong>on</strong> of u = x 2 which is itself a power functi<strong>on</strong> of x. Thus,dfdx = dex2dx = deu dudx du = (deu du )(du dx ) = eu (x 2 ) ′ = 2xe x2 ,where a substituti<strong>on</strong> u = x 2 was used to change f(x) into a true exp<strong>on</strong>ential functi<strong>on</strong> f(u).Therefore, the Chain Rule <strong>ca</strong>n be simply interpreted as the quotient between df and dx is equalto the quotient between df and du multiplied by the quotient between du and dx. Or simply,divide df/dx by du and then multiply it by du.However, if we simply regard x 2 as a functi<strong>on</strong> different from x, the actual substituti<strong>on</strong> u = x 2becomes unnecessary. Instead, the above derivative <strong>ca</strong>n be expressed asde x2dx = dx 2 dex2dx dx = 2 (dex2 dx 2 )(dx2 dx ) = (x 2 ) ′ = 2xe ex2 x2 .Generally, if f = f(g(x)) is a differentiable functi<strong>on</strong> of g and g is a differentiable functi<strong>on</strong> ofx, thendfdx = df dgdg dx ,where df, dg, and dx are, respectively, the differentials of the functi<strong>on</strong>s f, g, and x.Example: Calculate df/dx for f(x) = sin(ln(x 2 + e x )).Soluti<strong>on</strong>: f is a composite functi<strong>on</strong> of another composite functi<strong>on</strong>!15
dfdx = d sin(ln(x2 + e x ))d ln(x 2 + e x )d ln(x 2 + e x ) d(x 2 + e x )d(x 2 + e x ) dx= cos(ln(x 2 + e x 1))(x 2 + e x ) (2x + ex ).7 The Product Rule in Terms of DifferentialsThe Product Rule says if both u = u(x) and v = v(x) are differentiable functi<strong>on</strong>s of x, thend(uv)dx= v dudx + udv dx .Multiply both sides by the differential dx, we obtaind(uv) = vdu + udv,which is the Product Rule in terms of differentials.Example: Let u = x 2 and v = e sin(x2) , thend(uv) = vdu + udv = e sin(x2) d(x 2 ) + x 2 d(e sin(x2) )= 2xe sin(x2) dx + x 2 e sin(x2) cos(x 2 )(2x)dx = 2xe sin(x2) [1 + x 2 cos(x 2 )]dx.8 Other Properties of Differentials1. For any differentiable functi<strong>on</strong> F (x), dF = F ′ (x)dx (Re<strong>ca</strong>ll that F ′ (x) = dF/dx!).2. For any c<strong>on</strong>stant C, dC = 0.3. for any c<strong>on</strong>stant C and differentiable functi<strong>on</strong> F (x), d(CF ) = CdF = CF ′ (x)dx.4. For any differentiable functi<strong>on</strong>s u and v, d(u ± v) = du ± dv.16
9 The Fundamental Theorem in Terms of DifferentialsFundamental Theorem of <strong>Calculus</strong>: If F (x) is <strong>on</strong>e antiderivative of the functi<strong>on</strong> f(x), i.e.,F ′ (x) = f(x), then∫∫f(x)dx =∫F ′ (x)dx =dF = F (x) + C.Thus, the integral of the differential of a functi<strong>on</strong> F is equal to the functi<strong>on</strong> itselfplus an arbitrary c<strong>on</strong>stant. This is simply saying that differential and integral are inverse<strong>math</strong> operati<strong>on</strong>s of each other. If we first differentiate a functi<strong>on</strong> F (x) and then integrate thederivative F ′ (x) = f(x), we obtain F (x) itself plus an arbitrary c<strong>on</strong>stant. The opposite also istrue. If we first integrate a functi<strong>on</strong> f(x) and then differentiate the resulting integral F (x)+C,we obtain F ′ (x) = f(x) itself.Example:∫∫x 5 dx =d( x66 ) = x66 + C.Example:∫∫e −x dx =d(−e −x ) = −e −x + C.Example:∫∫cos(3x)dx =d( sin(3x) ) = sin(3x) + C.3 3Example:∫∫sec 2 xdx =dtanx = tanx + C.Example:∫∫cosh(3x)dx =d( sinh(3x) ) = sinh(3x) + C.3317
Example:∫∫11 + x dx = 2dtan −1 x = tan −1 + C.Example:∫∫11 − tanh 2 x dx =∫ 1 + cosh(2x)=dx = 1 ∫22∫1sech 2 x dx =cosh 2 xdxd(x + sinh(2x) ) = 1 sinh(2x)[x + ] + C.2 2 2For more indefinite integrals involving elementary functi<strong>on</strong>s, look at the first page of the tableof integrals provided at the end of the notes.10 Integrati<strong>on</strong> by Subsituti<strong>on</strong>Substituti<strong>on</strong> is a necessity when integrating a composite functi<strong>on</strong> since we <strong>ca</strong>nnot write downthe antiderivative of a composite functi<strong>on</strong> in a straightforward manner.Many students find it difficult to figure out the substituti<strong>on</strong> since for different functi<strong>on</strong>s thesubsituti<strong>on</strong>s are also different. However, there is a general rule in substituti<strong>on</strong>, namely, tochange the composite functi<strong>on</strong> into a simple, elementary functi<strong>on</strong>.Example: ∫ (sin( √ x)/ √ x)dx.Soluti<strong>on</strong>: Note that sin( √ x) is not an elementary sine functi<strong>on</strong> but a composite functi<strong>on</strong>. Thefirst goal in solving this integral is to change sin( √ x) into an elementary sine functi<strong>on</strong> throughsubstituti<strong>on</strong>. Once you realize this, u = √ x is an obvious subsituti<strong>on</strong>. Thus, du = u ′ dx =12 √ x dx, or dx = 2√ xdu = 2udu. Substitute into the integral, we obtain∫ sin(√ x)√ xdx =∫ sin(u)u∫2udu = 2∫sin(u)du = −2dcos(u) = −2cos(u)+C = −2cos( √ x)+C.Once you become more experienced with subsituti<strong>on</strong>s and differentials, you do not need to dothe actual substituti<strong>on</strong> but <strong>on</strong>ly symboli<strong>ca</strong>lly. Note that x = ( √ x) 2 ,18
∫ √ ∫ √ sin( x) sin( x)√ dx = √ d( √ ∫ √ sin( x)x) 2 = √ 2 √ xd √ ∫x = 2x x xsin( √ x)d √ x = −2cos( √ x)+C.Thus, as so<strong>on</strong> as you realize that √ x is the substituti<strong>on</strong>, your goal is to change the differentialin the integral dx into the differential of √ x which is d √ x.If you feel that you <strong>ca</strong>nnot do it without the actual substituti<strong>on</strong>, that is fine. You <strong>ca</strong>n alwaysdo the actual substituti<strong>on</strong>. I here simply want to teach you a way that actual subsituti<strong>on</strong> is nota necessity!Example: ∫ x 4 cos(x 5 )dx.Soluti<strong>on</strong>: Note that cos(x 5 ) is a composite functi<strong>on</strong> that becomes a simple cosine functi<strong>on</strong> <strong>on</strong>lyif the subsituti<strong>on</strong> u = x 5 is made. Since du = u ′ dx = 5x 4 dx, x 4 dx = 1 du. Thus,5Or alternatively,∫x 4 cos(x 5 )dx = 1 5∫cos(u)du = sin(u)5+ C = sin(x5 )5+ C.∫x 4 cos(x 5 )dx = 1 5∫cos(x 5 )dx 5 = 1 5 sin(x5 ) + C.Example: ∫ x √ x 2 + 1dx.Soluti<strong>on</strong>: Note that √ x 2 + 1 = (x 2 + 1) 1/2 is a composite functi<strong>on</strong>. We realize that u = x 2 + 1is a substituti<strong>on</strong>. du = u ′ dx = 2xdx implies xdx = 1 du. Thus,2Or alternatively,∫x √ x 2 + 1dx = 1 ∫ √udu 1 2 =22 3 u3/2 + C = (x2 + 1) 3/23+ C.∫x √ x 2 + 1dx = 1 2∫ √x2+ 1d(x 2 + 1) = 1 22 3 (x2 + 1) 3/2 + C = (x2 + 1) 3/2+ C.319
In many <strong>ca</strong>ses, substituti<strong>on</strong> is required even no obvious composite functi<strong>on</strong> is involved.Example: ∫ (ln x/x)dx (x > 0).Soluti<strong>on</strong>: The integrand ln x/x is not a composite functi<strong>on</strong>. Nevertheless, its antiderivative isnot obvious to <strong>ca</strong>lculate. We need to figure out that (1/x)dx = d ln x, thus by introducing thesubstituti<strong>on</strong> u = ln x, we obtained a differential of the functi<strong>on</strong> ln x which also appears in theintegrand. Therefore,∫∫(ln x/x)dx =udu = u22 + C = 1 2 (ln x)2 + C.It is more natural to c<strong>on</strong>sider this substituti<strong>on</strong> is an attempt to change the differential dx intosomething that is identi<strong>ca</strong>l to a functi<strong>on</strong> that appears in the integrand, namely d ln x. Thus,∫∫(ln x/x)dx =ln xd ln x = 1 2 (ln x)2 + C.Example: ∫ tan xdx.Soluti<strong>on</strong>: tan x is not a composite functi<strong>on</strong>. Nevertheless, it is not obvious to figure out tan xis the derivative of what functi<strong>on</strong>. However, if we write tan x = sin x/ cos x, we <strong>ca</strong>n regard1/ cos x as a composite functi<strong>on</strong>. We see that u = cosx is a <strong>ca</strong>ndidate for substituti<strong>on</strong> anddu = u ′ dx = −sin(x)dx. Thus,∫tanxdx =∫ sinxcosx dx = − ∫ duu= − ln |u| + C = − ln |cosx| + C.Or alternatively,∫tanxdx =∫ ∫ sinx dcosxcosx dx = − cosx= − ln |cosx| + C.Some substituti<strong>on</strong>s are standard in solving specific types of integrals.Example: Integrands of the type (a 2 − x 2 ) ±1/2 .In this <strong>ca</strong>se both x = asinu and x = acosu will be good. x = a tanh u also works (1−tanh 2 u =sech 2 u). Let’s pick x = asinu in this example. If you ask how <strong>ca</strong>n we find out that x = asinu20
is the substituti<strong>on</strong>, the answer is a 2 −x 2 = a 2 (1−sin 2 u) = a 2 cos 2 u. This will help us eliminatethe half power in the integrand. Note that with this substituti<strong>on</strong>, u = sin −1 (x/a), sinu = x/a,and cosu = √ 1 − x 2 /a 2 . Thus,∫∫dx√a2 − x = 2∫ ∫d(asinu) acosudu√a2 − a 2 sin 2 u = acosu=du = u + C = sin −1 x a + C.Note that ∫ dx/ √ 1 − x 2 = sin −1 x + C, we <strong>ca</strong>n make a simple substituti<strong>on</strong> x = au, thus∫∫dx√a2 − x = 2∫d(au)√a2 − a 2 u = 2du√1 − u2 = sin−1 u + C = sin −1 x a + C.Similarly,∫ √a2∫ √a2∫− x 2 dx = − a 2 sin 2 ud(asinu) = a 2∫ 1 + cos(2u)cos 2 udu = a 2 du2= a22sin(2u)[u + ] + C = a222 [u + sinucosu] + C = 1 2 [a2 sin −1 ( x a ) + x√ a 2 − x 2 ] + C.Example: Integrands of the type (a 2 + x 2 ) ±1/2 .x = asinh(u) is a good substituti<strong>on</strong> since a 2 + x 2 = a 2 + a 2 sinh 2 (u) = a 2 (1 + sinh 2 (u)) =a 2 cosh 2 (u), where the hyperbolic identity 1 + sinh 2 (u) = cosh 2 (u) was used. (x = a tan u isalso good since 1 + tan 2 u = sec 2 u!). Thus,∫∫dx√a2 + x = 2∫ ∫dasinh(u) acosh(u)du√a2 + a 2 sinh 2 (u) = =acosh(u)du = u + C = sinh −1 ( x a ) + C= ln |x + √ a 2 + x 2 | + C,where sinh −1 (x/a) = ln |x + √ a 2 + x 2 | − ln |a| was used. And,∫ √a2∫ √a2∫+ x 2 dx = + a 2 sinh 2 (u)dasinh(u) = a 221cosh 2 (u)du = a22∫[1 + cosh(2u)]du
= a2 sinh(2u)[u + ] + C = 1 2 22 [a2 sinh −1 ( x a ) + a2 sinh(u)cosh(u)] + C= 1 2 [a2 ln |x + √ a 2 + x 2 | + x √ a 2 + x 2 ] + C,where the following hyperbolic identities were used: sinh(2u) = 2sinh(u)cosh(u), sinh −1 (x/a) =ln |x + √ a 2 + x 2 | − ln |a|, and acosh(u) = √ a 2 + x 2 .Example: Integrands of the type (x 2 − a 2 ) ±1/2 .x = acosh(u) is prefered since x 2 − a 2 = a 2 scos 2 (u) − a 2 = a 2 [cosh 2 (u) − 1] = a 2 sinh 2 (u),where the hyperbolic identity cosh 2 (u)−1 = sinh 2 (u) was used. (x = a sec u is also good sincesec 2 u − 1 = tan 2 u!). Thus,∫∫dx√x2 − a = 2∫ ∫da cosh(u) a sinh(u)du==√a 2 cosh 2 (u) − a 2 a sinh(u)du = u + C= cosh −1 ( x a ) + C = ln |x + √ x 2 − a 2 | + C.Similarly∫ √x2− a 2 dx =∫ √a 2 cosh 2 (u) − a 2 da cosh(u) = a 2 ∫sinh 2 udu = a22∫[cosh(2u) − 1]du= 1 2 [x√ x 2 − a 2 − a 2 ln |x + √ x 2 − a 2 |] + C,where cosh u = x/a, sinh u = √ x 2 /a 2 − 1, and u = ln |x + √ x 2 − a 2 | − ln a were used.For more details <strong>on</strong> hyperbolic functi<strong>on</strong>s, read the last page of the table of integrals at the endof the notes.More Exercises <strong>on</strong> Substituti<strong>on</strong>:22
1. Substituti<strong>on</strong> aimed at eliminating a composite functi<strong>on</strong>Example:(i)∫∫ √ 4xd(2x 2 ∫32x 2 + 3 dx = √ + 3)32x 2 + 3 =u −1/3 du = 3 2 u2/3 + C = 3 2 (2x2 + 3) 2/3 + C.(ii)∫x√ dx = 1 ∫ 2x + 3 − 3√ d(2x + 3) = 1 ∫ u − 32x + 3 4 2x + 3 4 u du = 1 ∫1/2 4[u 1/2 − 3u −1/2 ]du= 1 4 [2 3 u3/2 − 6u 1/2 ] + C = 1 4 [2 3 (2x + 3)3/2 − 6(2x + 3) 1/2 ] + C.Edwards/Penney 5 th − ed 5.7 Problems.∫∫(1) (3x − 5) 17 dx (3)∫(7)∫(23)∫x sin(2x 2 )dx (9)x √ ∫2 − 3x 2 dx (31)x √ ∫x 2x 2 + 9dx (4) √ 32x 3 − 1 dx∫(1 − cosx) 5 dxsinxdx (15) √ dx 7x + 5∫cos 3 xsinxdx (33)tan 3 xsec 2 xdx(45)∫ cos(√ x)√ xdx (49)∫π/20(1 + 3sinθ) 3/2 cosθdθ (51)∫π/20e sinx cosxdxEdwards/Penney 5 th − ed 9.2 Problems.(7)∫(21)∫ √ √ ∫cot( y)csc( y)√ dy (11)ytan 4 (3x)sec 2 (3x)dx (25)∫ (ln t)e −cotx csc 2 1 0xdx (13) dt, (t > 0)t∫ √ (1 + x)4∫√ dx (31) x 2√ x + 2dx.x2. Substituti<strong>on</strong> to achieve a functi<strong>on</strong> in differential that appears in the integrand23
Example:(i) ∫e x ∫1 + e dx = 2xde x ∫1 + (e x ) = 2du1 + u 2 = tan−1 (u) + C = tan −1 (e x ) + C.(ii)∫x√e2x 2 − 1 dx = 1 ∫2dx 2√e2x 2 (1 − e −2x2 ) = 1 ∫2du√e2u(1 − e −2u ) = 1 ∫2due u√ 1 − (e −u ) 2= − 1 2∫de −u ∫√1 − (e−u) = −1 2 2dy√1 − y2 = 1 2 cos−1 (y)+C = 1 2 cos−1 (e −u )+C = 1 2 cos−1 (e −x2 )+C.Edwards/Penney 5 th − ed 9.2 Problems (difficult <strong>on</strong>es!).∫(17)sinθ)∫(27)e 2x∫1 + e dy, (u = 4x e2x ) (19)∫1(1 + t 2 )tan −1 t dt, (u = tan−1 t) (29)∫3x√ dx, (u = 1 − x4 x2 ) (23)1√e2x− 1 dx, (u = e−x )cosθdθ, (u =1 + sin 2 θ3. Special Trig<strong>on</strong>ometric Substituti<strong>on</strong>sExample:(i)∫x 3 ∫√ dx = 1 − x2sin 3 ∫ud sin u√1 − sin 2 u =∫sin 3 udu = −(1 − cos 2 u)d cos u= cos3 u3− cos u + C = (1 − x2 ) 3/23− √ 1 − x 2 + C.Edwards/Penney 5 th − ed 9.6 Problems (difficult <strong>on</strong>es!).∫∫ √ ∫1(1) √ dx, (x = 4 sin u) (9) x2 − 1dx, (x = cosh u) (11)16 − x2 x3 sinh u)x 3√ 9 + 4x 2 dx, (2x =24
∫ √ ∫1 − 4x2(13)dx, (2x = sin u) (19)x3 sinh u)x 2√1 + x2 dx, (x = sinh u) (27) ∫ √9+ 16x2 dx, (4x =11 Integrati<strong>on</strong> by PartsIntegrati<strong>on</strong> by Parts is the integral versi<strong>on</strong> of the Product Rule in differentiati<strong>on</strong>. The ProductRule in terms of differentials reads,Integrating both sides, we obtaind(uv) = vdu + udv.∫∫d(uv) =∫vdu +udv.Note that ∫ d(uv) = uv + C, the above equati<strong>on</strong> <strong>ca</strong>n be expressed in the following form,∫∫udv = uv −vdu.Generally speaking, we need to use Integrati<strong>on</strong> by Parts to solve many integrals that involvethe product between two functi<strong>on</strong>s. In many <strong>ca</strong>ses, Integrati<strong>on</strong> by Parts is most efficient insolving integrals of the product between a polynomial and an exp<strong>on</strong>ential, a logarithmic, or atrig<strong>on</strong>ometric functi<strong>on</strong>. It also applies to the product between exp<strong>on</strong>ential and trig<strong>on</strong>ometricfuncti<strong>on</strong>s.Example: ∫ xe x dx.Soluti<strong>on</strong>: In order to eliminate the power functi<strong>on</strong> x, we note that (x) ′ = 1. Thus,∫∫xe x dx =∫xde x = xe x −e x dx = xe x − e x + C.Example: ∫ x 2 cosxdx.Soluti<strong>on</strong>: In order to eliminate the power functi<strong>on</strong> x 2 , we note that (x 2 ) ′′ = 2. Thus, we needto use Integrati<strong>on</strong> by Patrs twice.25
∫∫x 2 cosxdx =∫x 2 dsinx = x 2 sinx −∫sinxdx 2 = x 2 sinx −2xd(−cosx)∫= x 2 sinx + 2∫xdcosx = x 2 sinx + 2xcosx − 2cosxdx = x 2 sinx + 2xcosx − 2sinx + C.Example: ∫ x 2 e −2x dx.Soluti<strong>on</strong>: In order to eliminate the power functi<strong>on</strong> x 2 , we note that (x 2 ) ′′ = 2. Thus, we needto use Integrati<strong>on</strong> by Patrs twice. However, the number (−2) <strong>ca</strong>n prove extremely annoyingand easily <strong>ca</strong>use errors. Here is how we use substituti<strong>on</strong> to avoid this problem.∫x 2 e −2x dx = −12 3 ∫(−2x) 2 e −2x d(−2x) = −18∫u 2 e u du = −1 ∫8 [u2 e u − 2ue u du]= −18 [u2 e u − 2(ue u − e u )] + C = −eu8 [u2 − 2u + 2] + C = −e−2x[4x 2 + 4x + 2] + C.8When integrating the product between a polynomial and a logarithmic functi<strong>on</strong>, the main goalis to eliminate the logarithmic functi<strong>on</strong> by differentiating it. This is be<strong>ca</strong>use (ln x) ′ = 1/x.Example: ∫ ln xdx, (x > 0).Soluti<strong>on</strong>:∫∫ln xdx = x ln x −∫xd ln x = x ln x −x dx x= x ln x − x + C.Example: ∫ x(ln x) 2 dx, (x > 0).Soluti<strong>on</strong>:∫x(ln x) 2 dx = 1 2∫(ln x) 2 dx 2 = 1 ∫2 [x2 (ln x) 2 −x 2 (2 ln x) dx x ] = 1 ∫2 [x2 (ln x) 2 −ln xdx 2 ]26
= 1 2 [x2 (ln x) 2 − x 2 ln x + x22 ] + C = x22 [(ln x)2 − ln x + 1 2 ] + C.More Exercises <strong>on</strong> Integrati<strong>on</strong> by Parts:Example:(i)∫∫t sin tdt = −∫td cos t = −[t cos t −cos tdt] = [sin t − t cos t] + C.(ii)∫∫tan −1 xdx = xtan −1 x −∫xdtan −1 x = xtan −1 x −x1 + x 2 dx= xtan −1 x − 1 2∫ d(1 + x 2 )1 + x 2 = xtan −1 x − ln √ 1 + x 2 + C.Edwards/Penney 5 th − ed 9.3 Problems.∫∫∫(1) xe 2x dx (5) x cos(3x)dx (7) x 3 ln xdx, (x > 0)∫ ∫∫√y(11) ln ydy (13) (ln t) 2 dt (19) csc 3 θdθ∫∫∫(21) x 2 tan −1 xdx (27) xcsc 2 xdx (19) csc 3 θdθ12 Integrati<strong>on</strong> by Partial Fracti<strong>on</strong>sRati<strong>on</strong>al functi<strong>on</strong>s are defined as the quotient between two polynomials:R(x) = P n(x)Q m (x)where P n (x) and Q m (x) are polynomials of degree n and m respectively. The method of partialfracti<strong>on</strong>s is an algebraic technique that decomposes R(x) into a sum of terms:27
R(x) = P n(x)Q m (x) = p(x) + F 1(x) + F 2 (x) + · · · F k (x),where p(x) is a polynomial and F i (x), (i = 1, 2, · · · , k) are fracti<strong>on</strong>s that <strong>ca</strong>n be integratedeasily.The method of partial fracti<strong>on</strong>s is an area many students find very difficult to learn. It isrelated to algebraic techniques that many students have not been trained to use. The mosttypi<strong>ca</strong>l claim is that there is no fixed formula to use. It is not our goal in this course to coverthis topics in great details (Read Edwards/Penney for more details). Here, we <strong>on</strong>ly study twosimple <strong>ca</strong>ses.Case I: Q m (x) is a power functi<strong>on</strong>, i.e., Q m (x) = (x − a) m (Q m (x) = x m if a = 0!).Example: ∫ 2x 2 − x + 3dx.This integral involves the simplest partial fracti<strong>on</strong>s:x 3A + B + CD= A D + B D + C D .Some may feel that it is easier to write the fracti<strong>on</strong>s in the following form: D −1 (A + B + C) =D −1 A + D −1 B + D −1 C. Thus,∫ 2x 2 ∫− x + 3dx =x 3[ 2x2x 3− x x 3 + 3 x 3 ]dx = ∫[ 2 x − 1 x 2 + 3 x 3 ]dx = ln x2 + 1 x − 32x 2 + C.Example:=∫ 2x 2 ∫− x + 3 2x 2 ∫x 2 − 2x + 1 dx = − x + 3 2(x − 1 + 1) 2 − (x − 1 + 1) + 3dx =dx(x − 1) 2 (x − 1) 2∫ 2(x − 1) 2 ∫+ 4(x − 1) + 2 − (x − 1) − 1 + 3 2(x − 1) 2 + 3(x − 1) + 4dx =dx(x − 1) 2 (x − 1) 2∫=[2 + 3x − 1 + 44]d(x − 1) = 2(x − 1) + 3 ln |x − 1| −(x − 1)2x − 1 + C.28
Case II: P n (x) = A is a c<strong>on</strong>stant and Q m (x) <strong>ca</strong>n be factorized into the form Q 2 (x) =(x − a)(x − b), Q 3 (x) = (x − a)(x − b)(x − c), or Q m (x) = (x − a 1 )(x − a 2 ) · · · (x − a m ).Example:∫I =∫2x 2 − 2x − 3 dx =∫2dx(x − 3)(x + 1) = 2A[x − 3 +Bx + 1 ]dx.ASince + B = 1implies that A(x + 1) + B(x − 3) = 1. Setting x = 3 in thisx−3 x+1 (x−3)(x+1)equati<strong>on</strong>, we obtain A = 1/4. Setting x = −1, we obtain B = −1/4. Thus,∫I = 2A[x − 3 +B ∫x + 1 ]dx = 2[ 1/4x − 3 + −1/4x + 1 ]dx = 1 ∫2dx[x − 3 −dxx + 1 ] = 1 − 3ln |x2 x + 1 | + C.Example:∫I =∫dx(x − 1)(x − 2)(x − 3) =A[x − 1 +Bx − 2 +Cx − 3 ]dx.SinceA + B + C = 1 implies that A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2) = 1.x−1 x−2 x−3Setting x = 1 in this equati<strong>on</strong>, we obtain A = 1/2. Setting x = 2, we obtain B = −1. Settingx = 3, we get C = 1/2. Thus,∫I =[ 1/2x − 1 + −1x − 2 + 1/2x − 3 ]dx = 1 |(x − 1)(x − 3)|ln + C.2 (x − 2) 2More Exercises <strong>on</strong> Integrati<strong>on</strong> by Partial Fracti<strong>on</strong>s:Examples:(i)∫∫dxx 2 − 3x =dxx(x − 3) = 1 ∫31[x − 3 − 1 x ]dx = 1 3ln |x− 3x | + C.29
(ii)∫∫dxx 2 + x − 6 =dx(x + 3)(x − 2) = 1 ∫51[x − 2 − 1x + 3 ]dx = 1 5ln |x− 2x + 3 | + C.(iii) ∫ ∫ ∫x − 1 x + 1 − 2x + 1 dx = x + 1dx =[1 − 2x + 1 ]dx = x − ln(x + 1)2 + C.Edwards/Penney 5 th − ed 9.5 Problems.∫ 2x 3 + 3x 2 ∫− 5dx(Li)dx (15)x 3 x 2 − 4∫x(Li)dx, (Hint : x = x + 2 − 2)x 2 + 5x + 6 (Li)∫(Li)∫(Li)∫x 2x 2 + x dx, (Hint : x2 = [(x + 1) − 1] 2 ) (23)dxx 2 − 2x − 8x 21 + x dx, (Hint : 2 x2 = 1 + x 2 − 1)∫x 2(x + 1) dx 313 Integrati<strong>on</strong> by Multiple TechniquesWhenever we encounter a complex integral, often the first thing to do is to use substituti<strong>on</strong> toeliminate the composite functi<strong>on</strong>(s).Example: I = ∫ x −3 e 1/x dx.Soluti<strong>on</strong>: Note that e 1/x is a composite fucnti<strong>on</strong>. The substituti<strong>on</strong> to change it into a simple,elementary functi<strong>on</strong> is u = 1/x, du = u ′ dx = −x −2 dx or x −2 dx = −du = d(−1/x). Thus,∫I =∫x −3 e 1/x dx =∫(1/x)e 1/x d(−1/x) = −ue u du = −[ue u − e u ] + C = e 1/x [1 − 1/x] + C.30
Example: I = ∫ cos( √ x)dx.Soluti<strong>on</strong>: Note that cos( √ x) is a composite fucnti<strong>on</strong>. The substituti<strong>on</strong> to change it into asimple, elementary functi<strong>on</strong> is u = √ x, du = u ′ dx = 12 √ dx or dx = 2√ xdu = 2udu. However,xnote that x = ( √ x) 2 ,∫I =cos( √ ∫x)dx =cos( √ x)d( √ ∫x) 2 =∫cos(u)du 2 = 2ucos(u)du∫= 2udsin(u) = 2[usin(u) + cos(u)] + C = 2[ √ xsin( √ x) + cos( √ x)] + C.Example: I = ∫ sec(x)dx = ∫ (1/cos(x))dx.Soluti<strong>on</strong>: This is a difficult integral. Note that 1/cos(x) is a composite fucnti<strong>on</strong>. One substituti<strong>on</strong>to change it into a simple, elementary functi<strong>on</strong> is u = cos(x). However, x = cos −1 (u) anddx = x ′ du = √ −du1−u 2. 1/ √ 1 − u 2 is still a composite functi<strong>on</strong> that is difficult to deal with. Hereis how we <strong>ca</strong>n deal with it. It involves trig<strong>on</strong>ometric substituti<strong>on</strong> followed by partial fracti<strong>on</strong>s.∫I =∫ ∫dx cos(x)dxcos(x) = cos 2 (x) =∫dsin(x)1 − sin 2 (x) =∫du1 − u = − 21[u − 1 − 1u + 1 ]du 2= ln√| u + 1u − 1 | + C = ln | u + 1+ 1√ | + C = ln |sin(x) | + C = ln |tan(x) + sec(x)| + C,1 − u2 cos(x)where u = sin(x) and √ 1 − u 2 = cos(x) were used!More Exercises <strong>on</strong> Integrati<strong>on</strong> by Multiple techniques:1. Substituti<strong>on</strong> followed by Integrati<strong>on</strong> by PartsExample:(i)∫x 3 sin(x 2 )dx = 1 2∫x 2 sin(x 2 )dx 2 = 1 2∫u sin udu = −12∫ud cos u= 1 2 [sin u − u cos u] + C = 1 2 [sin(x2 ) − x 2 cos(x 2 )] + C.31
(ii)∫xtan −1√ ∫xdx =( √ x) 2 tan −1√ xd( √ ∫x) 2 =u 2 tan −1 udu 2 = 1 2∫tan −1 udu 4= 1 ∫2 [u4 tan −1 u−u 4 dtan −1 u] = 1 ∫ u2 [u4 tan −1 4 − 1 + 1u−du] = 1 ∫1 + u 2 2 [u4 tan −1 u−(u 2 −1+ 11 + u 2 )du]= 1 2 [u4 tan −1 u − u33 + u − tan−1 u] + C = 1 2 [(x2 − 1)tan −1√ x − x3/23 + √ x] + C.Edwards/Penney 5 th − ed 9.3 Problems.∫(15)∫(23)∫(31)x √ ∫x + 3dx, u = x + 3 (17)sec −1√ xdx, u = √ ∫x (29)ln xx √ x dx, u = √ ∫x (37)x 5√ x 3 + 1dx, u = x 3 + 1∫x 3 cos(x 2 )dx, u = x 2 (25)e −√x dx, u = √ xtan −1√ xdx, u = √ x2. Substituti<strong>on</strong> followed by Integrati<strong>on</strong> by Partial Fracti<strong>on</strong>sExample:(i)∫cos xdxsin 2 x − sin x − 6 = ∫∫d sin xsin 2 x − sin x − 6 =du(u − 3)(u + 2) = 1 − 3ln |u5 u + 2 |+C = 1 x − 3ln |sin5 sin x + 2 |+CEdwards/Penney 5 th − ed 9.5 Problems (diffifult <strong>on</strong>es!)∫(40)sec 2 ttan 3 t + tan 2 t dt(37) ∫e 4t∫(e 2t − 1) dt (39) 31 + ln tt(3 + 2 ln t) 2 dt2. Integrati<strong>on</strong> by Parts followed by Partial Fracti<strong>on</strong>s32
Example:(i)∫∫(2x + 2) tan −1 xdx =∫tan −1 xd(x 2 + 2x) = (x 2 + 2x) tan −1 x −(x 2 + 2x)d tan −1 x∫ 1 + x= (x 2 + 2x) tan −1 2 ∫+ 2x − 1x −dx = (x 2 + 2x) tan −1 x −1 + x 2[dx + d(1 + x2 )1 + x 2 − dx1 + x 2 ]= (x 2 + 2x) tan −1 x − x − ln(1 + x 2 ) + tan −1 x + C = (x + 1) 2 tan −1 x − x − ln(1 + x 2 ) + C.Li’s Problems (diffifult <strong>on</strong>es!)∫(Li)(2t + 1) ln tdt33
14 Appli<strong>ca</strong>ti<strong>on</strong>s of <strong>Integral</strong>sThe central idea in all sorts of appli<strong>ca</strong>ti<strong>on</strong> problems involving integrals is breaking the wholeinto pieces and then adding the pieces up to obtain the whole.14.1 Area under a curve.We know the area of a rectangle is:Area = height × width,(see Fig. 1(a)). This formula <strong>ca</strong>n not be applied to the area under the curve in Fig. 1(b)be<strong>ca</strong>use the top side of the area is NOT a straight, horiz<strong>on</strong>tal line. However, if we divide thearea into infinitely many rectangles with infinitely thin width dx, the infinitely small area ofthe rectangle lo<strong>ca</strong>ted at x isdA = height × infintely small width = f(x)dx.Adding up the area of all the thin rectangles, we obtain the area under the curveArea = A(b) − A(a) =∫A(b)∫ bdA = f(x)dx,0awhere A(x) = ∫ xf(s)ds is the area under the curve between a and x.a14.2 Volume of a solid.We know the volume of a cylinder isV olume = (cross − secti<strong>on</strong> area) × length,(see Fig. 2(a)). This formula does not apply to the volume of the solid obtained by rotatinga curve around the x-axis (see Fig. 2(b)). This is be<strong>ca</strong>use the cross-secti<strong>on</strong> area is NOT a34
¢¢¢¢¢¢¢¢¢¢¢¢c<strong>on</strong>stant but changes al<strong>on</strong>g the length of this “cylinder”. However, if we cut this “cylinder”into infinitely many disks of infinitely small thickness dx, the volume of such a disk lo<strong>ca</strong>ted atx isdV = cross − secti<strong>on</strong> area × infintely small thickness = πr 2 dx = π(f(x)) 2 dx.Adding up the volume of all such thin disks, we obtain the volume of the solidV olume = V (b) − V (a) =∫V (b)∫ bdV = π (f(x)) 2 dx,0awhere V (x) = π ∫ xa (f(s))2 ds is the volume of the “cylinder” between a and x.14.3 Arc length.We know the length of the hypotenuse of a right triangle ishypotenuse = √ base 2 + height 2 ,(see Fig. 3(a)). This formula does not apply to the length of the curve in Fig. 3(b) be<strong>ca</strong>use thecurve is NOT a straight line and the geometric shape of the area enclosed in solid lines is NOTA=hwhy=f(x)¡¡¡¡¡w(a)a¡¡¡¡¡¡dx(b)xbFigure 1: Area under a curve.¡35
Lry=f(x)¢¡¢¢¡¢¢¡¢¢¡¢¢¡¢¢¡¢¢¡¢¢¡¢¢¡¢x¢¡¢¢¡¢¢¡¢V= πr 2 L(a)a¢¡¢¢¡¢¢¡¢¢¡¢¢¡¢dx(b)Figure 2: Volume of a solid obtained by rotating a curve.ba right triangle. However, if we “cut” the curve <strong>on</strong> interval [a, b] into infinitely many segmentseach with a horiz<strong>on</strong>tal width dx, each curve segment <strong>ca</strong>n be regarded as the hypotenuse of aninifinitely small right triangle with base dx and height dy. Thus, the length of <strong>on</strong>e such curvesegments lo<strong>ca</strong>ted at x isdl = √ (infinitely small base) 2 + (infinitely small height) 2= √ dx 2 + dy 2 =√[1 + ( dydx )2 ]dx 2 = [ √ 1 + (y ′ ) 2 ]dx.The total length of this curve is obtained by adding the lengths of all the segments together.L =∫ l(b)0dl =∫ ba[ √ 1 + (y ′ ) 2 ]dx,where l(x) = ∫ xa [√ 1 + [y ′ (s)] 2 ]ds is the length of the curve between a and x.Example: Calculate the length of the curve y = x 2 /2 <strong>on</strong> the interval [−2, 2].Soluti<strong>on</strong>: Be<strong>ca</strong>use of the symmetry, we <strong>on</strong>ly need to <strong>ca</strong>lculate the length <strong>on</strong> the interval [0, 2]and multiply it by two. Thus, noticing that y ′ = x,∫ l(2)L = 2 dl = 2l(0)∫ 20[ √ 1 + (y ′ ) 2 ]dx = 2∫ 20[ √ 1 + x 2 ]dx = 2F (x)| 2 0,36
y=f(x)height=hhyp2= b2+ h2base=b(a)a2 2 2dl = dx + dydydx(b)bFigure 3: Length of a curve.xwhere F (x) = ∫ [ √ 1 + x 2 ]dx. Now, this integral <strong>ca</strong>n be solved by the standard substituti<strong>on</strong>x = sinh(u). Note that 1 + x 2 = 1 + sinh 2 (u) = cosh 2 (u) and dx = dsinh(u) = cosh(u)du, weobtain by subsituti<strong>on</strong>∫F (x) =[ √ ∫1 + x 2 ]dx =∫cosh(u)cosh(u)du =cosh 2 (u)du = 1 2∫[1 + cosh(2u)]du= 1 2 [u + 1 2 sinh(2u)] + C = 1 2 [u + sinh(u)cosh(u)] + C = 1 2 [ln |x + √ 1 + x 2 | + x √ 1 + x 2 ] + C,where the inverse substituti<strong>on</strong> sinh(u) = x, cosh(u) = √ 1 + x 2 , and u = sinh −1 (x) = ln |x +√1 + x2 | were used in the last step.Thus,L = 2F (x)| 2 0 = ln |2 + √ 1 + 2 2 | + 2 √ 1 + 2 2 = ln(2 + √ 5) + 2 √ 5 ≈ 5.916.14.4 From density to mass.We have learned that the mass c<strong>on</strong>tained in a solid isMass = density × volume.37
The formula is no l<strong>on</strong>ger valid if the density is NOT a c<strong>on</strong>stant but is n<strong>on</strong>uniformly distributedin the solid. However, if you cut the mass up into infinitely many pieces of inifinitely smallvolume dV , then the density in this little volume <strong>ca</strong>n be c<strong>on</strong>sidered a c<strong>on</strong>stant so that we <strong>ca</strong>napply the above formula. Thus, the infinitely small mass c<strong>on</strong>tained in the little volume isdm = density × (infinitely small volume) = ρdV.The total mass is obtained by adding up the masses of all such small pieces.m =∫ m(V (b))0dm =∫ V (b)V (a)ρdV,where V (x) = ∫ V (x)0dV is the volume of the porti<strong>on</strong> of the solid between a and x; whilem(V (x)) = ∫ m(V (x))0dm is the mass c<strong>on</strong>tained in the volume V (x).Example: Calculate the total amount of pollutant in an exhaust pipe filled with pollutedliquid which c<strong>on</strong>nects a factory to a river. The pipe is 100 m l<strong>on</strong>g with a diameter of 1 m.The density of the pollutant in the pipe is ρ(x) = e −x/10 (kg/m 3 ), x is the distance in metersfrom the factory.Soluti<strong>on</strong>: Since the density <strong>on</strong>ly varies as a functi<strong>on</strong> of the distance x, we <strong>ca</strong>n “cut” the pipeinto thin cylindri<strong>ca</strong>l disks al<strong>on</strong>g the axis of the pipe. Thus, dV = d(πr 2 x) = πr 2 dx. Using theintegral form of the formula, we obtain=∫ 1000m =ρ(x)πr 2 dx = π 4∫ m(100)m(0)∫ 1000dm =∫ V (100)V (0)ρ(x)dVe −x/10 dx = 10π4 [1 − e−10 ] ≈ 7.85 (kg).Example: The air density h meters above the earth’s surface is ρ(h) = 1.28e −0.000124h (kg/m 3 ).Find the mass of a cylindri<strong>ca</strong>l column of air 4 meters in diameter and 25 kilometers high. (3points)Soluti<strong>on</strong>: Similar to the previous problem, the density <strong>on</strong>ly varies as a functi<strong>on</strong> of the altitutdeh. Thus, we “cut” this air column into horiz<strong>on</strong>tal slices of thin disks with volume dV = πr 2 dh.38
∫ m(25,000) ∫ V (25,000) ∫ 25,000∫ 25,000m = dm = ρdV = ρ(h)πr 2 dh = ρ(h)π2 2 dhm(0)V (0)00= 4 × 1.28π−0.000124 e−0.000124h | 25,0000 = 4 × 1.28π0.000124 [1 − e−3.1 ] = 123, 873.71 (kg).Very often the density is given as mass per unit area (two dimensi<strong>on</strong>al density). In this <strong>ca</strong>se,dm = ρdA, where dA is an infinitely small area. Similarly, if the density is given as mass perunit length (<strong>on</strong>e dimensi<strong>on</strong>al density), then dm = ρdx, where dx is an infinitely small length.Example: The density of bacteria growing in a circular col<strong>on</strong>y of radius 1 cm is observed tobe ρ(r) = 1 − r 2 , 0 ≤ r ≤ 1 (in units of <strong>on</strong>e milli<strong>on</strong> cells per square centimeter), where r isthe distance (in cm) from the center of the col<strong>on</strong>y. What is the total number of bacteria inthe col<strong>on</strong>y?Soluti<strong>on</strong>: This is a problem involving a two dimensi<strong>on</strong>al density. However, the density varies<strong>on</strong>ly in the radial directi<strong>on</strong>. Thus, we <strong>ca</strong>n divide the circular col<strong>on</strong>y into infinitely many c<strong>on</strong>centricring areas. We <strong>ca</strong>n use the formula (# of cells in a ring) = density × (area of the ring),i.e., dm = ρ(r)dA. The area, dA, of a ring of bacteria col<strong>on</strong>y with radius r and width dr isdA = π(r + dr) 2 − πr 2 = 2πrdr + πdr 2 = 2πrdr, where dr 2 is infinitely smaller than dr, thus isignored as dr → 0. Another way to <strong>ca</strong>lculate the area of the ring is to c<strong>on</strong>sider the ring as a rectangulararea of length 2πr (i.e., the circumference) and width dr, dA = length×width = 2πrdr.Or simply, A = πr 2 , thus dA = d(πr 2 ) = πdr 2 = 2πdr.m =∫ m(1)m(0)= π∫ 10dm =∫ A(1)A(0)ρdA =∫ 10ρ(r)d(πr 2 ) =∫ 10ρ(r)2πrdr(1 − r 2 )dr 2 = −π (1 − r2 ) 2| 1 02= π 2 (milli<strong>on</strong>s).Example: The density of a band of protein al<strong>on</strong>g a <strong>on</strong>e-dimensi<strong>on</strong>al strip of gel in an electrophoresisexperiment is given by ρ(x) = 6(x−1)(2−x) for 1 ≤ x ≤ 2, where x is the distanceal<strong>on</strong>g the strip in cm and ρ(x) is the protein density (i.e. protein mass per cm) at distance x.Find the total mass of the protein in the band for 1 ≤ x ≤ 2.Soluti<strong>on</strong>: This is a problem involving a <strong>on</strong>e dimensi<strong>on</strong>al density. Thus, (infinitely small mass) =density × (infinitely small length), i.e., dm = ρ(x)dx.39
∫ m(2) ∫ 2m = dm = ρ(x)dx = −6m(1)1∫ 21(x−1)(x−2)dx = −[2x 3 −9x 2 +12x]∣21= −4−(−5) = 1.14.5 Work d<strong>on</strong>e by a varying force.Elementary physics tells us that work d<strong>on</strong>e by a force isW ork = force × distance.The formula is no l<strong>on</strong>ger valid if the force is NOT a c<strong>on</strong>stant over the distance covered butchanges. However, if we “cut” the distance into infinitely many segments of inifinitely shortdistances dx, then the force <strong>on</strong> that little distance <strong>ca</strong>n be c<strong>on</strong>sidered a c<strong>on</strong>stant. Thus, theinfinitely small work d<strong>on</strong>e <strong>on</strong> a distance dx isdW = force × (infintely small distance) = fdx.The total work is obtained by adding up the work d<strong>on</strong>e <strong>on</strong> all such small distances.W =∫ W (b)W (a)dW =∫ bafdx.Example: Calculate total work d<strong>on</strong>e by the earth’s gravitati<strong>on</strong>al force <strong>on</strong> a satellite of massm when it is being launched into the space. This is the lower limit of energy required to launchit into space. Note that the gravitati<strong>on</strong>al force experienced by a mass m lo<strong>ca</strong>ted at a distancer from the center of the earth is: f(r) = GM e m/r 2 , where G is the universal gravitati<strong>on</strong>alc<strong>on</strong>stant and M e the mass of the earth. Assume that the earth is a sphere with radius R e andthat r = ∞ when the satellite is “outside” the earth’s gravitati<strong>on</strong>al field. (C<strong>on</strong>sider G, M e ,R e , and m are known c<strong>on</strong>stants.)Soluti<strong>on</strong>: We know that work d<strong>on</strong>e <strong>on</strong> an infinitely small distance dr when the satellite islo<strong>ca</strong>ted at a distance r from the center of the earth is: dW = f(r)dr = (GM e m/r 2 )dr. Thus,W =∫ W (∞)W (R e)dW =∫ ∞R e∫ ∞1f(r)dr = GM e mR er dr. 240
<strong>Integral</strong>s that involve a limit which is ∞ are <strong>ca</strong>lled improper integrals. We shall study improperintegrals in more detail later. In many <strong>ca</strong>ses, we <strong>ca</strong>n just treat ∞ as a normal upper limit ofthis definite integral.∫ ∞1W = GM e mR er dr = GM em(− 1 2 r )|∞ R e= GM em= mgR e ,R ewhere the gravitai<strong>on</strong>al c<strong>on</strong>stant <strong>on</strong> the surface of the earth g = GM e /R 2 ewas used.Example: For the time interval 0 ≤ t ≤ 10, write down the definite integral determining thetotal work d<strong>on</strong>e by a force, f(t) (Newt<strong>on</strong>), experienced by a particle moving <strong>on</strong> a straightline with a speed, v(t) > 0 (m/s). f(t) and v(t) always point to the same directi<strong>on</strong> and bothchange with time t (s).Soluti<strong>on</strong>:W =∫ W (10)W (0)dW =∫ x(10)x(0)f(t)dx =∫ 100f(t)( dxdt )dt = ∫ 100f(t)v(t)dt.41
TABLE OF INTEGRALS15 Elementary integralsAll of these follow immediately from the table of derivatives. They should be memorized.∫∫∫∫∫cf(x) dx = cc dx = cx + Cf(x) dx∫∫∫[αf + βg] dx = αdx = x + C∫x r dx = xr+11r + 1 + C (r ≠ −1) dx = ln |x| + C∫ xe x dx = e x + Cln x dx = x ln x − x + C∫f dx + βg dxThe following are elementary integrals involving trig<strong>on</strong>ometric functi<strong>on</strong>s.∫∫∫∫∫sin x dx = − cos x + Csec 2 x dx = tan x + C1a 2 + x 2 dx = 1 a tan−1 x a + Ctan x dx = − ln | cos x| + Csec x dx = ln | tan x + sec x| + C∫∫∫∫∫cos x dx = sin x + Ccsc 2 x dx = − cot x + C1√a2 − x 2 dx = sin−1 x a + Ccot x dx = ln | sin x| + Ccsc x dx = ln | cot x − csc x| + CThe following are elementary integrals involving hyperbolic functi<strong>on</strong>s (for details read theSecti<strong>on</strong> <strong>on</strong> Hyperbolic Functi<strong>on</strong>s).∫∫∫∫∫sinh x dx = cosh x + Csech 2 x dx = tanh x + C1a 2 − x 2 dx = 1 a tanh−1 x a + Ctanh x dx = ln(cosh x) + Csech x dx = tan −1 (sinh x) + C∫∫∫∫∫cosh x dx = sinh x + Ccsch 2 x dx = − coth x + C1√a2 + x 2 dx = sinh−1 x a + Ccoth x dx = ln | sinh x| + Ccsch x dx = ln | coth x − csch x| + C42
16 A selecti<strong>on</strong> of more compli<strong>ca</strong>ted integralsThese begin with the two basic formulas, change of variables and integrati<strong>on</strong> by parts.∫∫f(g(x))g ′ (x) dx = f(u) du where u = g(x) (change of variables)∫∫∫∫∫∫∫∫∫f(g(x)) dx =e cx dx = 1 c ecx + C (c ≠ 0)f(u) dx du where u = g(x) (different form of the same change of variables)dua x dx = 1ln a ax + C (for a > 0, a ≠ 1)ln x dx = x ln x − x + C1x 2 + a 2 dx = 1 a arctan x a + C, a ≠ 01x 2 − a dx = 1 ∣ ∣∣ 2 2a ln x − ax + a∣ +C, a ≠ 01√a2 − x 2 dx = arcsin x a + C, a > 01√x2 ± a 2 dx = ln |x + √ x 2 ± a 2 | + C∫To compute1dx we complete the squarex 2 + bx + cx 2 + bx + c = x 2 + bx + b24 + c − b24 = (x + b 2) 2+ c − b24If c − b 2 /4 > 0, set it equal to a 2 ; if < 0 equal to −a 2 ; and if = 0 forget it. In any event youwill arrive after the change of variables u = x + b at <strong>on</strong>e of the three integrals2∫∫∫1u 2 + a du, 11 2 u 2 − a du, 2 u du 2∫ √x2± a 2 dx = 1 2(x √ x 2 ± a 2 ± a 2 ln∣ x + √ x 2 ± a 2 ∣ ∣∣)+ C∫x n e cx dx = x n ecxc − n ∫x n−1 e cx dx etc. This is to be used repetaedly until you arrive at thec<strong>ca</strong>se n = 0, which you <strong>ca</strong>n do easily.43
17 Hyperbolic Functi<strong>on</strong>sDefiniti<strong>on</strong>: The most frequently used hyperbolic functi<strong>on</strong>s are defined by,sinh x = ex − e −x2cosh x = ex + e −x2tanh x = sinh xcosh x = ex − e −xe x + e −xwhere sinh x and tanh x are odd like sin x and tan x, while cosh x is even like cos x (see Fig. 4for plots of these functi<strong>on</strong>s). The definiti<strong>on</strong>s of other hyperbolic functi<strong>on</strong>s are identi<strong>ca</strong>l to thecorresp<strong>on</strong>ding trig functi<strong>on</strong>s.10−4 0 4−10sinh(x)63−3 −2 −1 0 1 2 3cosh(x)1−4 −2 0 2 4−1tanh(x)Figure 4: Graphs of sinh x, cosh x, and tanh x.We <strong>ca</strong>n easily derive all the corresp<strong>on</strong>ding identities for hyperbolic functi<strong>on</strong>s by using thedefiniti<strong>on</strong>. However, it helps us memorize these identities if we know the relati<strong>on</strong>ship betweensinh x, cosh x and sin x, cos x in complex analysis.Combining sinh x = ex − e −x2and sin x = eix − e −ix2i, we obtain sinh(x) = −i sin(ix).Combining cosh x = ex + e −x2and cos x = eix + e −ix2, we obtain cosh(x) = cos(ix).where the c<strong>on</strong>stant i = √ −1 and i 2 = −1. Now, we <strong>ca</strong>n write down these identities:cos 2 x + sin 2 x = 1 cosh 2 x − sinh 2 x = 11 + tan 2 x = sec 2 x 1 − tanh 2 x = sech 2 xcot 2 x + 1 = csc 2 xcoth 2 x − 1 = csch 2 xcos 2 x = (1 + cos 2x)/2cosh 2 x = (1 + cosh 2x)/2sin 2 x = 1 − cos 2 x = (1 − cos 2x)/2sinh 2 x = cosh 2 x − 1 = (−1 + cosh 2x)/2sin 2x = 2 sin x cos xsinh 2x = 2 sinh x cosh xsin(x ± y) = sin x cos y ± cos x sin ysinh(x ± y) = sinh x cosh y ± cosh x sinh ycos(x ± y) = cos x cos y ∓ sin x sin ycosh(x ± y) = cosh x cosh y ± sinh x sinh y44
The Advantage of Hyperbolic Functi<strong>on</strong>sThe advantage of hyperbolic functi<strong>on</strong>s is that their inverse functi<strong>on</strong>s <strong>ca</strong>n be expressed in termsof elementary functi<strong>on</strong>s. We here show how to solve for the inverse hyperbolic functi<strong>on</strong>s.By definiti<strong>on</strong>, u = sinh x = ex − e −x.2Multiplying both sides by 2e x , we obtain2ue x = (e x ) 2 − 1 which simplifies to (e x ) 2 − 2u(e x ) − 1 = 0.This is a quadratic in e x , Thus,e x = [2u ± √ (2u) 2 + 4]/2 = u ± √ u 2 + 1.The negative sign in ± does not make sense since e x > 0 for all x.Thus, x = ln e x = ln |u + √ u 2 + 1| which impliesx = sinh −1 u = ln |u + √ u 2 + 1|.Similarly, by definiti<strong>on</strong>, u = cosh x = ex + e −x.2We <strong>ca</strong>n solve this equati<strong>on</strong> for the two branches of the inverse:x = cosh −1 u = ln |u ± √ u 2 − 1| = ± ln |u + √ u 2 − 1|.Similarly, foru = tanh x = ex − e −xe x + e −x .We <strong>ca</strong>n solve the equati<strong>on</strong> for the inverse:x = tanh −1 u = 1 2ln |1+ u1 − u |. 45
The Advantage of Trig<strong>on</strong>ometric Functi<strong>on</strong>sRelati<strong>on</strong>s between different trig<strong>on</strong>ometric functi<strong>on</strong>s are very important since when we differentiateor integrate <strong>on</strong>e trig functi<strong>on</strong> we obtain another trig functi<strong>on</strong>. When we use trigsubstituti<strong>on</strong>, it is often a necessity for us to know the definiti<strong>on</strong> of other trig functi<strong>on</strong>s that isrelated to the <strong>on</strong>e we use in the substituti<strong>on</strong>.Example: Calculate the derivative of y = sin −1 x.Soluti<strong>on</strong>: y = sin −1 x means sin y = x. Differentiate both sides of sin y = x, we obtaincos yy ′ = 1 ⇒ y ′ = 1cos y .In order to express y ′ in terms of x, we need to express cos y in terms of x. Given that sin y = x,we <strong>ca</strong>n obtain cos y = √ cos 2 y = √ 1 − sin 2 y = √ 1 − x 2 by using trig identities. However, itis easier to c<strong>on</strong>struct a right triangle with an angle y. The opposite side must be x while thehypotenuse must be 1, thus the adjacent side is √ 1 − x 2 . Therefore,cos y =√adjacent side 1 − xhypotenuse = 21= √ 1 − x 2 .Example: Calculate the integral ∫ √ 1 + x 2 dx.Soluti<strong>on</strong>: This integral requires standard trig substituti<strong>on</strong> x = tan u or x = sinh u. Let’s usex = sinh u. Re<strong>ca</strong>ll that 1 + sinh 2 u = cosh 2 u and that dx = d sinh u = cosh udu,∫ √1∫ √ ∫+ x2 dx = 1 + sinh 2 ud sinh u =cosh 2 udu= 1 2∫[1 + cosh(2u)]du = 1 2 [u + 1 2 sinh(2u)] + C = 1 [u + sinh u cosh u] + C.2where hyperbolic identities cosh 2 u = [1 + cosh(2u)]/2 and sinh(2u) = 2 sinh u cosh u wereused. However, we need to express the soluti<strong>on</strong> in terms of x. Since x = sinh u was thesubstituti<strong>on</strong>, we know right away u = sinh −1 x = ln |x + √ 1 + x 2 | and sinh u = x, but how toexpress cosh u in terms of x? We <strong>ca</strong>n solve it using hyperbolic identities. cosh u = √ √cosh 2 u =1 + sinh 2 u = √ 1 + x 2 . Thus,12 [u + sinh u cosh u] = 1 2 [ln |x + √ 1 + x 2 | + x √ 1 + x 2 ].46
However, the following table shows how <strong>ca</strong>n we exploit the similarity between trig and hyperbolicfuncti<strong>on</strong>s to find out the relati<strong>on</strong>ship between different hyperbolic functi<strong>on</strong>s.Relati<strong>on</strong>s between different trig and between different hyperbolic functi<strong>on</strong>s=====================================================Given <strong>on</strong>e trig func, sin x = u = u 1Given sinh x = −i sin(ix) = u, sin(ix) = iu = iu 1Draw a right triangleDraw the corresp<strong>on</strong>ding right trianglewith an angle x, opp = u, hyp = 1: with an angle ix, opp = iu, hyp = 1:u1x1 − u 2iu1ix1 − ( i u) 2 =21 + uThus, other trig functi<strong>on</strong>s arecos x = adjhyp = √ 1 − u 2tan x = oppadj =u√1 − u2By formal analogy,cosh x = cos(ix) = adjhyp = √ 1 + u 2tanh x = −i tan(ix) = −i oppadj =u√1 + u2=====================================================47
=====================================================Given that, cos x = u = u 1Given cosh x = cos(ix) = u = u 1Draw a right triangleDraw the corresp<strong>on</strong>ding right trianglewith an angle x, adj = u, hyp = 1: with an angle ix, adj = iu, hyp = 1:1 − u 2 1ux1 − u 2 = i u2− 1(u = cosh(x) > 1)u1ixThus, other trig functi<strong>on</strong>s aresin x = opphyp = √ 1 − u 2tan x = opp√1 − uadj = 2uBy formal analogy,sinh x = −i sin(ix) = −i opphyp = √ u 2 − 1tanh x = −i tan(ix) = −i opp√adj = u2 − 1u=====================================================Given that, tan x = u = u 1Given tanh x = (−i) tan(ix) = u, tan(ix) = iu = iu 1Draw a right triangleDraw the corresp<strong>on</strong>ding right trianglewith an angle x, opp = u, adj = 1: with an angle ix, opp = iu, adj = 1:21 + uThus, we know all other trig functi<strong>on</strong>susin x = opp xhyp = u√1 1 + u2cos x = adjhyp = 1√1 + u2By formal analogy,sinh x = −i sin(ix) = −i opphyp =u√1 − u2cosh x = cos(ix) = adjhyp = 1√1 − u2=====================================================48
21 + ( i u)iu1=ix1 − u 249