Solução_Calculo_Stewart_6e

F.
TX.10
7 TECHNIQUES OF INTEGRATION
7.1 Integration by Parts
1. Let u =lnx, dv = x 2 dx ⇒ du = 1 dx, v = 1 x 3 x3 .ThenbyEquation2,
x 2 ln xdx=(lnx) 1
x3 − 1
x3
1
3 3 x dx =
1
3 x3 ln x − 1 3 x 2 dx = 1 3 x3 ln x − 1 1 x3 + C
3 3
= 1 3 x3 ln x − 1 9 x3 + C or 1 x3
ln x − 1 3 3 + C
Note: A mnemonic device which is helpful for selecting u when using integration by parts is the LIATE principle of precedence for u:
Logarithmic
Inverse trigonometric
Algebraic
Trigonometric
Exponential
If the integrand has several factors, then we try to choose among them a u which appears as high as possible on the list. For example, in xe 2x dx
the integrand is xe 2x , which is the product of an algebraic function (x) and an exponential function (e 2x ). Since Algebraic appears before Exponential,
we choose u = x. Sometimes the integration turns out to be similar regardless of the selection of u and dv, but it is advisable to refer to LIATE when in
doubt.
3. Let u = x, dv =cos5xdx ⇒ du = dx, v = 1 5
sin 5x. ThenbyEquation2,
x cos 5xdx=
1
x sin 5x − 1
sin 5xdx= 1 1
5 5 5
x sin 5x +
25
cos 5x + C.
5. Let u = r, dv = e r/2 dr ⇒ du = dr, v =2e r/2 .Then re r/2 dr =2re r/2 − 2e r/2 dr =2re r/2 − 4e r/2 + C.
7. Let u = x 2 , dv =sinπxdx ⇒ du =2xdxand v = − 1 π
cos πx. Then
I =
x 2 sin πxdx = − 1 π x2 cos πx + 2 π x cos πxdx (). Next let U = x, dV =cosπx dx ⇒ dU = dx,
V = 1 sin πx, so π
x cos πxdx = 1 x sin πx − 1
π π sin πx dx =
1
1
π
x sin πx + cos πx + C
π 1.
2
Substituting for x cos πx dx in (), we get
I = − 1 π x2 cos πx + 2 1
1
x sin πx + cos πx + C
π π π 2 1 = −
1
π x2 cos πx + 2 x sin πx + 2 cos πx + C,whereC = 2 C π 2 π 3 π 1.
9. Let u =ln(2x +1), dv = dx ⇒ du = 2 dx, v = x. Then
2x +1
2x
(2x +1)− 1
ln(2x +1)dx = x ln(2x +1)−
2x +1 dx = x ln(2x +1)− dx
2x +1
= x ln(2x +1)− 1 − 1
dx = x ln(2x +1)− x + 1 ln(2x +1)+C
2
2x +1
11. Let u =arctan4t, dv = dt ⇒ du =
arctan 4tdt= t arctan 4t −
= 1 2
(2x +1)ln(2x +1)− x + C
4
1+(4t) dt = 4
dt, v = t. Then
2 1+16t2 4t
1+16t 2 dt = t arctan 4t − 1 8
32t
1+16t 2 dt = t arctan 4t − 1 8 ln(1 + 16t2 )+C.
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