MATH221 Problem set 1: Review of integration by parts and by substitution
(Solutions)
Do the following exercises from Calculus, 5th ed. by Hughes-Hallett–Gleason–McCallum–et
al.:
7.1 (pg. 338) 3, 13, 24, 29, 30.
7.2 (pg. 346) 9, 4, 20.
Ch. 7 review (pg. 385) 33, 41, 22, 25, 26.
Just for fun If the preceding are too easy, the following integral should be more fun:
Z
Find
sec x dx.
(Trick: Multiply the integrand by
sec x+tan x
sec x+tan x
and make a substitution).
7.1
3
Z
2
tet dt.
Take u = t2 , so that du = 2t dt. We have
Z
Z
2
1
1 2
tet dt =
eu du = et + C.
2
2
13
Z
t2 (t3 − 3)10 dt.
Take u = (t3 − 3), so that du = 3t2 dt. We have
Z
Z
1
1 3
t2 (t3 − 3)10 dt =
u10 du =
(t − 3)11 + C.
3
33
24
Z √
cos 3t sin 3t dt.
Take u = cos 3t, so that du = −3 sin 3t dt. Then
Z √
Z
1 √
2
cos 3t sin 3t dt = −
u du = − (cos 3t)3/2 + C.
3
9
29
Z
Take u = ln z, so that du =
Z
(ln z)2
dz.
z
dz
z .
We have
Z
(ln z)2
1
dz = u2 du = (ln z)3 + C.
z
3
1
30
Z
et + 1
dt.
et + t
Take u = et + t, so that du = (et + 1) dt. Then
Z
Z t
du
e +1
dt
=
= ln et + t + C.
t
e +t
u
7.2
9
Z
y ln y dy.
Let u(y) = ln y, v 0 (y) = y, so that u0 (y) = y1 , v(y) =
Z
y 2 ln y
y ln y dy =
−
2
4
Z
Z
y2
2 .
Integrating by parts,
y
y 2 ln y y 2
dy =
−
+ C.
2
2
4
t2 sin t dt.
Integrate by parts twice. First, let u(t) = t2 , v 0 (t) = sin t. Then u(t) = 2t, v(t) = − cos t
and we have
Z
Z
t2 sin t dt = −t2 cos t + 2 t cos t dt.
Now, to find the last integral, take u(t) = t, v 0 (t) = cos t, so that u0 (t) = 1, v(t) = sin t
and integrate by parts to get
Z
Z
t cos t dt = t sin t − sin t dt = t sin t + cos t + C.
Combining the above,
Z
20
t2 sin t dt = −t2 cos t + 2t sin t + 2 cos t + C.
Z
ln x
dx.
x2
Integrate by parts. Take u(x) = ln x, v 0 (x) = x12 , so that u0 (x) = x1 , v(x) = − x1 . We have
Z
Z
ln x
dx
ln x + 1
ln x
ln x 1
+
− +C =−
+ C.
=
−
=−
2
2
x
x
x
x
x
x
2
7 Review
33
Z
cos2 θ dθ.
, we have
Using the standard identity cos2 θ = cos 2θ+1
2
Z
Z
Z
Z
cos 2θ + 1
sin 2θ θ
cos 2θ
1
cos2 θ dθ =
dθ =
dθ +
dθ =
+ + C.
2
2
2
4
2
41
Z
tan θ dθ.
We have tan θ =
sin θ
cos θ ,
so that, making the substitution u = cos θ, du = − sin θ dθ,
Z
Z
du
tan θ dθ = −
= − ln | cos θ| + C.
u
22
Z
√
x 1 − x dx.
√
Let u(x) = x, v 0 (x) = 1 − x, so that u0 (x) = 1, v(x) = − 23 (1 − x)3/2 . Integrating by
parts, we have
Z
Z
√
2
2
(1 − x)3/2 dx.
x 1 − x dx = − x(1 − x)3/2 +
3
3
2
4
= − x(1 − x)3/2 − (1 − x)5/2 + C.
3
15
25
Z
(ln x)2 dx.
Do the following somewhat tricky integration by parts: take u(x) = (ln x)2 , v 0 (x) = 1, so
x
that u0 (x) = 2 ln
x , v(x) = x. We have
Z
Z
2
2
(ln x) dx = x(ln x) − 2 ln x dx.
To find the last integral, let u(x) = ln x, v 0 (x) = 1, so that u(x) = x1 , v 0 (x) = x. Then
Z
Z
ln x dx = x ln x − dx = x ln x − x + C.
Combining the above,
Z
26
(ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C.
Z
ln(x2 ) dx.
2
Take u(x) = ln(x2 ), v 0 (x) = 1, so that u0 (x) = 2x
x2 = x , v(x) = x. We have
Z
Z
ln(x2 ) dx = x ln(x2 ) − 2 dx = x ln(x2 ) − 2x + C.
3
Just for fun
Using the hint, we the integrand becomes
Z
sec x + tan x
sec x
dx.
sec x + tan x
Let u(x) = sec x + tan x. Then u0 (x) = (sec x tan x + sec2 x) (check!), so that
Z
Z
Z
sec x + tan x
sec2 x + sec x tan x
du
sec x
dx =
dx =
= ln | sec x + tan x| + C.
sec x + tan x
sec x + tan x
u
4