INTEGRATION BY PARTS
237
28.50
28.51
28.52
28.53
28.54
28.55
28.56
28.57
Find fe^dx.
First make the substitution x = w , dx = 2w dw. Then J e * dx = 2 / we
w dw. By the reduction formula in Problem 28.42, / we
w dw = we
w - J e" dw = we" — e" — e
w (w — 1). So we obtain 2e
w (w — 1) +
C = 2e
vT (v*-l) + C.
Evaluate / Jt(ox + ft)
3 dx by integration by parts.
Then J x(ox + ft)
3
Do Problem 28.51 by means of a substitution.
The region under the curve y = cos x between x = 0 and x = ir/2 is revolved about the y-axis. Find the
volume of the resulting solid.
By the cylindrical shell formula.
cos x dx. By the solution of Problem 19.33,
with the aid of Problem 28.6
Find / (sin'
1 x)
2 dx
Make the substitution y = sin ' x,
Then
/ (sin"
1 x)
2 dx = J y
2 yi -sin
2 y dy =
Find f x" In x dx for n ^ — 1.
Then
Let w = In x, dv — x" dx,
Derive the reduction formula
Then
Find J Jt
5 (ln *)
2 dx.
Let u = (In x)
n , .dv = x
m dx, du
By the reduction formula of Problem 28.56, J *
5 (ln x)
2
Let u = x, dv = (ax + ft)
3 dx, du = dx,
\5ax -(ax + b)] + C
(ax + ft)
5 + C
(ax + ft)
4 dx
(4ax - ft) + C.
Let
u = ax + b,
du = a dx.
Then
J x(ax + ft)
3
(J M" du - / ftw
3 (4ox - ft) + C.
/ y
2 cos y dy = (y
2 - 2) sin y + 2y cos j> + C = [(sin"
1 *)
2 - 2] x + 2(sin
-1 x)Vl - x
2 + C.
we obtain 2ir(x sin x + cos x) ]„* = 2ir[(Trl2 - 0) - (0 + 1)] = 2-n(-nl2 - 1) = TT(TT - 2).
In
In
In