CHAPTER 28
Integration by Parts
In Problems 28.1-28.24, find the indicated antiderivative.
28.1
28.2
28.3
28.4
28.5
28.6
28.7
28.8
232
We use integration by parts again for the latter integral: let u = cos x, dv = e* dx, du = -sin x dx, v = e*.
Then
J e' cos x dx = e' cos x + J e* sin x dx.
Substituting in (1),
J e* sin x dx = e* sin x - (e* cos * +
J e* sin * dx) - e' sin * - e' cos x - J e* sin A: dx. Thus, 2 J e* sin x dx = e* (sin x - cos x) + C, J e
v sin x dx
from which je*(sin AT - cos x) + C,.
JxV'dx.
We use integration by parts: fudv = uv — fv du. In this case, let M = x
2 , dv = e * dx. Then dw =
2xdx, v = ~e~*. Hence, J x
2 e~* dx - -x
2 e~* + 2 J xe~* dx. [To calculate the latter, we use another
integration by parts: u = x, dv = e~* dx; du = dx, v = —e~*. Then J xe~' dx = — xe~* + J e~" dx =
-xe~' - e'" = -e~"(x + 1).] Hence, J x
2 e~' dx = -x
2 e~* + 2[-e~"(x + 1)] + C = -e~"(x
2 + 2x + 2) + C.
/ e' sin x dx.
Let M = sin x, dv = e' dx, du = cos x dx, v = e*. Then
Let M=je
3 , dv = e"dx, du = 3x
2 , v = e
x . Then J xV gives, with x replaced by-x, J xV dx = c*(x
2 -2x + 2) + C. Hence, J xV dx = e'(x
3 ~ ^ + 6x - 6) + C.
/ xV dx.
/sin ' x dx
Let
w = sin ' x,
dv = dx,
du = (l/Vl -x
2 ) dx,
i; = x.
Then
Jsin
1 xdx = xsin 'x(x/Vl - x
2 ) dx = x sin '
T^7 + c.
(l-x
2 )~"
2 (-2x)dx = xsin^
1
2(l-x
2 )"
2 + C = xsnT'x +
J x sin x dx.
Let w = x, du=sinxdx, du = dx, v = -cosx. Then J xsin xdx = —xcosx + Jcosx dx =-xcosx +
sin x + C.
J x
2 cos x dx.
Let
w = x
2
, dv = cosxdx,
dw = 2xdx,
u = sinx.
Then, using Problem 28.5,
Jx
2 cosxdx =
x
2 sin x — 2 J x sin x dx = x
2 sin x - 2(—x cos x + sin x) + C = (x
2 — 2) sin x + 2x cos x + C.
| cos (In x) dx.
Let
x = e
y ~"
/2 ,
cos (In x) = sin y,
dx = e
y ~"
12 dy,
and use Problem 28.2:
J cos (In x) dx =
e""
2 J e
y sin y dy = e~"
2 [^e"(sm y - cos y)} + C = ^x[cos (In x) + sin (In x)] + C.
f x cos (5x — 1) dx.
Let M=X, dv =cos(5x — 1) dx, du = dx,
sin (5x — 1). Then Jxcos(5x—1)
e* sin x dx = e" sin x - e* cos x dx.
(1)