Integration by Parts solved questions

CHAPTER 28
Integration by Parts

In Problems 28.1-28.24, find the indicated antiderivative.

28.1 JxV'dx.
We use integration by parts: fudv = uv — f v du. In this case, let M = x2, dv = e * dx. Then dw =
2xdx, v = ~e~*. Hence, J x2e~* dx - -x2e~* + 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 x2e~' dx = -x2e~* + 2[-e~"(x + 1)] + C = -e~"(x2 + 2x + 2) + C.

28.2 / e' sin x dx.
Let M = sin x, dv = e' dx, du = cos x dx, v = e*. Then

e* sin x dx = e" sin x - e* cos x dx. (1)

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,.

28.3 / xV dx.
Let M=je 3 , dv = e"dx, du = 3x2, v = ex. Then J xV <fc = *V - 3 J *V dx. But Problem 28.1
gives, with x replaced by-x, J xV dx = c*(x2 -2x + 2) + C. Hence, J xV dx = e'(x3 ~ ^ + 6x - 6) + C.


28.4 /sin ' x dx
Let w = sin ' x, dv = dx, du = ( l / V l -x 2 ) dx, i; = x. Then Jsin 1
x d x = xsin ' x -
2 2 2 1 2 2
(x/Vl - x ) dx = x sin ' (l-x )~" (-2x)dx = xsin^ 2(l-x )" + C = xsnT'x +
T^7 + c.

28.5 J x sin x dx.
Let w = x, d u = s i n x d x , du = dx, v = -cosx. Then J xsin xdx = —xcosx + Jcosx dx =-xcosx +
sin x + C.

28.6 J x2 cos x dx.
Let w = x2, dv = cosxdx, dw = 2xdx, u = sinx. Then, using Problem 28.5, Jx 2 cosxdx =
x sin x — 2 J x sin x dx = x 2 sin x - 2(—x cos x + sin x) + C = (x2 — 2) sin x + 2x cos x + C.
2



28.7 | cos (In x) dx.
Let x = e y ~" /2 , cos (In x) = sin y, dx = ey~"12 dy, and use Problem 28.2: J cos (In x) dx =
2 y 2
e"" J e sin y dy = e~" [^e"(sm y - cos y)} + C = ^x[cos (In x) + sin (In x)] + C.

28.8 f x cos (5x — 1) dx.
Let M = X , dv =cos(5x — 1) dx, du = dx, sin (5x — 1). Then Jxcos(5x—1)



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28.9 J e"' cos fee <&.
Let M = cos fee, dv = e""dx, du=-bsinbx, v = (\la)e". Then J e" cos bxdx = (\la)e cos bx +
(b/a) I e°* sin bx dx. Apply integration by parts to the latter: a = sin bx, dv = e°* dx, du = b cos bx,
v = (\ld)e°*. So / e" sin bxdx = (l/a)ea" sin bx - (b/«) JV* cos bxdx. Hence, by substitution,
J e°* cos bx dx = (l/a)e'"'cosbx + (b/a)[(l/a)e'"smbx-(b/a)$ e°*cosbxdx] = (l/a)e" cos fee +
(bla2)eax sin fee - (62/a2) J e* cos fee dx. Thus, (1 + b2/a2) J eaf cos fee dx = (e"/a2)(a cos bx + b sin fee) + C,
f e" cos bx dx = [e"/(a2 +fc"Wo cos fcx + b sin fee) + C,.

28.10 / sin2 x dx.
Let w = sinx, du = sinxdx, du=cosxdx, u = —cosx. Then J sin2 x dx = -sin xcosx + J cos2 x<it =
-sin x cos jc + J (1 - sin2 ;c) dx = —sin x cos * 4- AT - J sin2 x dx. So 2 J sin2 JT dx = x — sin jr cos x + C,
2
f sin x dx= 5 (x - sin jc cos x) + C,.

28.11 f cos3 x dx.
J cos3 je dx = J cos jc (1 - sin2 *) dc = J cos .* dx — / sin2 x cos x dx = sin sin3 x + C.

28.12 | cos4 x dx.

/ cos4 A: (1 + 2 cos 2x + cos2 2x)



28.13 $xe3* dx.

Let M = AT, rfu = e31 dx, du — dx, Then



28.14 J A: sec2 x dx.
Let u = x, dv = sec je dx, da = dx, v = tan x. Then J x sec2 x dx = x tan x - J tan x dx = x tan x —
Inlsecxl + C.

28.15 J je cos2 x dx.

Let M = x, dv = cos2 A: etc, du = dx, Then J" x cos2 x dx =
(2 sin 2x +
cos 2*) + C.

28.16 J (In x)2 dr.
Let x = e" and use Problem 28.1: J (Inx) 2 dx = -/ fV dt = e~'(t2 + 2t + 2) + C = x[(lnx) 2 -
2 In x + 2] + C.

28.17 J x sin 2x dx.

Let 2* = y and use Problem 28.5: / * sin 2x dx y sin y dy (-y cos y + sin y) + C =
(-2x cos 2x + sin 2*) + C.

28.18 J x sin (x2) dx.

Use a substitution M = x2, du = 2x dx. Then / x sin x2 sin u du = cos u + C =
cos x + C.