1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
integration by parts, In the above rule, there are two terms on R.H.S. and in both the terms integral of the second function is involve. Therefore in the product of two functions if one of the two functions is not directly integrable (e.g. log x, sin –1 x, cos –1 x, tan –1 x etc.) we take it as the first function and the remaining function is taken as the second function. If there is no other function, then unity is taken as the second function. If in the integral both the functions are easily integrable, then the first function is chosen in such a way that the derivative of the function is a simple functions and the function thus obtained under the integral sign is easily integrable than the original function. 2. ∫ sin( bx + c) dx e ax = = a e 2 ax a + b e ax 2 2 + b 3. ∫ cos( bx + c) dx e ax = = a e 2 ax a + b e ax 2 2 + b 4. ∫ ekx {kf(x) + f '(x)} dx = e kx f(x) [a sin (bx + c) – b cos (bx + c)] 2 ⎡ −1 b ⎤ sin ⎢bx + c − tan ⎥ ⎣ a ⎦ [a cos (bx + c) + b sin(bx + c)] 2 ⎡ −1 b ⎤ cos ⎢bx + c − tan ⎥ ⎣ a ⎦ 5. ⎛ x ⎞ ∫ log e x = x(logex – 1) = x loge ⎜ ⎟ ⎝ e ⎠ Integration of Trigonometric Functions : 1. To evaluate the integrals of the form I = ∫ sin m x cos n x dx, where m and n are rational numbers. (i) Substitute sin x = t, if n is odd; (ii) Substitute cos x = t, if m is odd; (iii) Substitute tan x = t, if m + n is a negative even integer; and 1 1 (iv) Substitute cot x = t, if (m + 1) + (n – 1) is an 2 2 integer. 2. Integrals of the form ∫ R (sin x, cos x) dx, where R is a rational function of sin x and cos x, are transformed into integrals of a rational function by the substitution x tan = t, where –π < x < π. This is the so called 2 universal substitution. Sometimes it is more x convenient to make the substitution cot = t for 2 0 < x < 2π. The above substitution enables us to integrate any function of the form R (sin x, cos x). However, in practice, it sometimes leads to extremely complex rational functions. In some cases, the integral can be simplified by – (i) Substituting sin x = t, if the integral is of the form ∫ R (sin x) cos x dx. (ii) Substituting cos x = t, if the integral is of the form ∫ R (cos x) sin x dx. dt (iii) Substituting tan x = t, i.e. dx = , if the 2 1+ t integral is dependent only on tan x. Some Useful Integrals : dx 1. (When a > b) ∫ a + b cos x 2 = tan 2 2 a − b –1 ⎡ − ⎤ ⎢ tan ⎥ ⎢⎣ + 2⎥⎦ x a b + c a b dx 2. (When a < b) ∫ a + b cos x x b − a tan − a + b 1 = – log a 2 2 b − a x b − a tan + a + b a dx 3. (when a = b) ∫ a + b cos x = 1 x tan + c a 2 dx 4. (When a > b) ∫ a + bsin x XtraEdge for IIT-JEE 44 FEBRUARY 2012 = 2 2 a − b 2 tan –1 dx 5. (When a < b) ∫ a + bsin x = 1 2 b − a 2 log ⎧ ⎛ x ⎞ ⎫ ⎪a tan⎜ ⎟ + b ⎪ ⎪ ⎝ 2 ⎠ ⎪ ⎨ ⎬ + c 2 2 ⎪ a − b ⎪ ⎪⎩ ⎪⎭ ⎛ x ⎞ a tan⎜ ⎟ + b − ⎝ 2 ⎠ ⎛ x ⎞ a tan⎜ ⎟ + b + ⎝ 2 ⎠ b b 2 2 − a − a dx 6. (When a = b) ∫ a + bsin x = 1 [tan x – sec x] + c a 2 2 + c
MATHS Functions with their Periods : Function Period sin (ax + b), cos (ax + b), sec (ax + b), cosec (ax + b) 2π/a tan(ax + b), cot (ax + b) π/a |sin (ax + b)|, |cos (ax + b)|, |sec (ax + b)|, |cosec (ax + b)| π/a |tan (ax + b)|, |cot (ax + b)| π/2a Trigonometrical Equations with their General Solution: Trgonometrical equation General Solution sin θ = 0 θ = nπ cos θ = 0 θ = nπ + π/2 tan θ = 0 θ = nπ sin θ = 1 θ = 2nπ + π/2 cos θ = 1 θ = 2nπ sin θ = sin α θ = nπ + (–1) n α cos θ = cos α θ = 2nπ ± α tan θ = tan α θ = nπ + α sin 2 θ = sin 2 α θ = nπ ± α tan 2 θ = tan 2 α θ = nπ ± α cos 2 θ = cos 2 α θ = nπ ± α sin θ = sin α * cosθ = cosα sin θ = sin α * tan θ = tan α tan θ = tan α * cosθ = cosα TRIGONOMETRICAL EQUATION θ = 2nπ + α θ = 2nπ + α θ = 2nπ + α * If α be the least positive value of θ which satisfy two given trigonometrical equations, then the general value of θ will be 2nπ + α. Note : 1. If while solving an equation we have to square it, then the roots found after squaring must be checked whether they satisfy the original equation or not. e.g. Let x = 3. Squaring, we get x 2 = 9, ∴ x = 3 and – 3 but x = – 3 does not satisfy the original equation x = 3. 2. Any value of x which makes both R.H.S. and L.H.S. equal will be a root but the value of x for which ∞ = ∞ will not be a solution as it is an indeterminate form. 3. If xy = xz, then x(y – z) = 0 ⇒ either x = 0 or y z y = z or both. But = ⇒ y = z only and not x x x = 0, as it will make ∞ = ∞. Similarly, if ay = az, then it will also imply y = z only as a ≠ 0 being a constant. Similarly, x + y = x + z ⇒ y = z and x – y = x – z ⇒ y = z. Here we do not take x = 0 as in the above because x is an additive factor and not multiplicative factor. 4. When cos θ = 0, then sin θ = 1 or –1. We have to verify which value of sin θ is to be chosen which ⎛ 1 ⎞ satisfies the equation cos θ = 0 ⇒ θ = ⎜n + ⎟ π ⎝ 2 ⎠ If sin θ = 1, then obviously n = even. But if sin θ = –1, then n = odd. Similarly, when sin θ = 0, then θ = nπ and cos θ = 1 or –1. If cos θ = 1, then n is even and if cos θ = –1, then n is odd. 5. The equations a cos θ ± b sin θ = c are solved as follows : Put a = r cos α, b = r sin α so that r = and α = tan –1 b/a. Mathematics Fundamentals XtraEdge for IIT-JEE 45 FEBRUARY 2012 2 a + b The given equation becomes r[cos θ cos α ± sin θ sin α] = c ; c c cos (θ ± α) = provided ≤ 1. r r Relation between the sides and the angle of a triangle: 1. Sine formula : 2
- Page 3 and 4: Volume - 7 Issue - 8 February, 2012
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integration by parts, In the above rule, there are two<br />
terms on R.H.S. and in both the terms integral of the<br />
second function is involve. Therefore in the product<br />
of two functions if one of the two functions is not<br />
directly integrable (e.g. log x, sin –1 x, cos –1 x, tan –1 x<br />
etc.) we take it as the first function and the remaining<br />
function is taken as the second function. If there is no<br />
other function, then unity is taken as the second<br />
function. If in the integral both the functions are<br />
easily integrable, then the first function is chosen in<br />
such a way that the derivative of the function is a<br />
simple functions and the function thus obtained under<br />
the integral sign is easily integrable than the original<br />
function.<br />
2. ∫ sin( bx + c)<br />
dx<br />
e ax<br />
=<br />
=<br />
a<br />
e<br />
2<br />
ax<br />
a<br />
+ b<br />
e ax<br />
2<br />
2<br />
+ b<br />
3. ∫ cos( bx + c)<br />
dx<br />
e ax<br />
=<br />
=<br />
a<br />
e<br />
2<br />
ax<br />
a<br />
+ b<br />
e ax<br />
2<br />
2<br />
+ b<br />
4. ∫ ekx {kf(x) + f '(x)} dx = e kx f(x)<br />
[a sin (bx + c) – b cos (bx + c)]<br />
2<br />
⎡<br />
−1<br />
b ⎤<br />
sin ⎢bx<br />
+ c − tan ⎥<br />
⎣<br />
a ⎦<br />
[a cos (bx + c) + b sin(bx + c)]<br />
2<br />
⎡<br />
−1<br />
b ⎤<br />
cos ⎢bx<br />
+ c − tan ⎥<br />
⎣<br />
a ⎦<br />
5.<br />
⎛ x ⎞<br />
∫ log e x = x(logex – 1) = x loge ⎜ ⎟<br />
⎝ e ⎠<br />
Integration of Trigonometric Functions :<br />
<strong>1.</strong> To evaluate the integrals of the form<br />
I = ∫ sin m x cos n x dx, where m and n are rational<br />
numbers.<br />
(i) Substitute sin x = t, if n is odd;<br />
(ii) Substitute cos x = t, if m is odd;<br />
(iii) Substitute tan x = t, if m + n is a negative even<br />
integer; and<br />
1 1<br />
(iv) Substitute cot x = t, if (m + 1) + (n – 1) is an<br />
2 2<br />
integer.<br />
2. Integrals of the form ∫ R (sin x, cos x) dx, where R is<br />
a rational function of sin x and cos x, are transformed<br />
into integrals of a rational function by the substitution<br />
x<br />
tan = t, where –π < x < π. This is the so called<br />
2<br />
universal substitution. Sometimes it is more<br />
x<br />
convenient to make the substitution cot = t for<br />
2<br />
0 < x < 2π.<br />
The above substitution enables us to integrate any<br />
function of the form R (sin x, cos x). However, in<br />
practice, it sometimes leads to extremely complex<br />
rational functions. In some cases, the integral can be<br />
simplified by –<br />
(i) Substituting sin x = t, if the integral is of the form<br />
∫ R (sin x) cos x dx.<br />
(ii) Substituting cos x = t, if the integral is of the form<br />
∫ R (cos x) sin x dx.<br />
dt<br />
(iii) Substituting tan x = t, i.e. dx = , if the<br />
2<br />
1+ t<br />
integral is dependent only on tan x.<br />
Some Useful Integrals :<br />
dx<br />
<strong>1.</strong> (When a > b) ∫ a + b cos x<br />
2<br />
= tan<br />
2 2<br />
a − b<br />
–1 ⎡ − ⎤<br />
⎢ tan ⎥<br />
⎢⎣<br />
+ 2⎥⎦<br />
x a b<br />
+ c<br />
a b<br />
dx<br />
2. (When a < b) ∫ a + b cos x<br />
x<br />
b − a tan − a + b<br />
1<br />
= – log a<br />
2 2<br />
b − a<br />
x<br />
b − a tan + a + b<br />
a<br />
dx<br />
3. (when a = b) ∫ a + b cos x<br />
= 1 x<br />
tan + c<br />
a 2<br />
dx<br />
4. (When a > b) ∫ a + bsin<br />
x<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 44 FEBRUARY <strong>2012</strong><br />
=<br />
2<br />
2<br />
a − b<br />
2<br />
tan –1<br />
dx<br />
5. (When a < b) ∫ a + bsin<br />
x<br />
=<br />
1<br />
2<br />
b − a<br />
2<br />
log<br />
⎧ ⎛ x ⎞ ⎫<br />
⎪a<br />
tan⎜<br />
⎟ + b ⎪<br />
⎪ ⎝ 2 ⎠ ⎪<br />
⎨<br />
⎬ + c<br />
2 2<br />
⎪ a − b ⎪<br />
⎪⎩<br />
⎪⎭<br />
⎛ x ⎞<br />
a tan⎜<br />
⎟ + b −<br />
⎝ 2 ⎠<br />
⎛ x ⎞<br />
a tan⎜<br />
⎟ + b +<br />
⎝ 2 ⎠<br />
b<br />
b<br />
2<br />
2<br />
− a<br />
− a<br />
dx<br />
6. (When a = b) ∫ a + bsin<br />
x<br />
= 1<br />
[tan x – sec x] + c<br />
a<br />
2<br />
2<br />
+ c