MATH 105 921
Solutions to Integration Exercises
MATH 105 921 Solutions to Integration Exercises
s2 + 1
ds
s2 − 1
Z
1)
Solution: Performing polynomial long division, we have that:
Z 2
Z
s +1
2
ds
=
(1
+
) ds
s2 − 1
s2 − 1
Z
Z
2
ds
=
ds +
2
s −1
Z
2
=s+
ds
2
s −1
Using partial fraction on the remaining integral, we get:
s2
A
B
A(s + 1) + B(s − 1)
(A + B)s + (A − B)
2
=
+
=
=
−1
s−1 s+1
(s + 1)(s − 1)
s2 − 1
Thus, A + B = 0 and A − B = 2. Adding the two equations together yields 2A = 2,
that is, A = 1, and B = −1. So, we have that:
Z
Z
Z
1
1
2
ds =
ds −
ds
2
s −1
s−1
s+1
Therefore,
Z
0
Z
2)
s2 + 1
ds = s +
s2 − 1
Z
2
ds
s2 − 1
Z
Z
1
1
=s+
ds −
ds
s−1
s+1
= s + ln |s − 1| − ln |s + 1| + C
√
x 1 + 2x dx
4
Solution: Using direct substitution with u = 1 + 2x and du = 2dx, we may write
Page 1 of 22
MATH 105 921
Solutions to Integration Exercises
x = 12 (u − 1). Moreover, when x = 4, u = 9, and when x = 0, u = 1. Thus,
Z
0
√
x 1 + 2x dx =
4
Z
3)
1
√
1
(u − 1) u du
4
Z9 1
1
1 3
(u 2 − u 2 ) du
=
9 4
1 5 1 3
= ( u 2 − u 2 ) |19
10
6
1
243 27
1
− )
=( − )−(
10 6
10
6
−298
=
15
Z
sin2 x cos2 x dx
Solution: Using half-angle identities
Z
Z
2
2
sin x cos x dx =
Z
=
Z
=
sin2 x =
1−cos(2x)
2
and cos2 x =
1+cos(2x)
,
2
we get:
1
(1 − cos(2x))(1 + cos(2x)) dx
4
1
(1 − cos2 (2x)) dx
4
Z
1
1
dx −
cos2 (2x) dx
4
4
Z
x 1
cos2 (2x) dx
= −
4 4
On the remaining integral, we apply the half-angle identity cos2 (2x) =
obtain:
Z
Z
x 1
1 + cos(4x)
2
dx = + sin(4x) + C
cos (2x) dx =
2
2 8
1+cos(4x)
,
2
Hence,
Z
Z
4)
sin2 x cos2 x dx =
x 1 x 1
x
1
− ( + sin(4x)) + C = −
sin(4x) + C
4 4 2 8
8 32
√
sin( w) dw
Page 2 of 22
and
MATH 105 921
Solutions to Integration Exercises
√
Solution: Using direct substitution with t = w, and dt =
√
2 w dt = 2t dt, we get:
Z
Z
√
sin( w) dw = 2t sin t dt
1
√
dw,
2 w
that is, dw =
Using integration by part method with u = 2t and dv = sin t dt, so du = 2 dt and
v = − cos t, we get:
Z
Z
2t sin t dt = −2t cos t + 2 cos t dt = −2t cos t + 2 sin t + C
Therefore,
Z
Z
5)
√
√
√
√
sin( w) dw = −2 w cos( w) + 2 sin( w) + C
ln(x)
dx
x
Solution: Using direct substitution with u = ln(x) and du =
1
x
dx, we get:
Z
ln(x)
u2
dx = u du =
+C
x
2
Z
ln(x)
1
⇒
dx = (ln(x))2 + C
x
2
Z
Z
6)
sin t cos(2t) dt
Solution: Recall the double-angle formula that cos(2t) = 2 cos2 t − 1, we get:
Z
Z
sin t cos(2t) dt = sin t(2 cos2 t − 1) dt
Z
Z
Z
2
= 2 sin t cos t dt − sin t dt = 2 sin t cos2 t dt + cos t
On the remaining integral, using direct substitution with u = cos t and du = − sin t dt,
we have that:
Z
Z
2
2
2
2 sin t cos t dt = −2u2 du = − u3 + C = − cos3 t + C
3
3
Page 3 of 22
MATH 105 921
Solutions to Integration Exercises
Therefore,
Z
Z
7)
2
sin t cos(2t) dt = − cos3 t + cos t + C
3
x+1
dx
4 + x2
Solution: Observe that we may split the integral as follows:
Z
Z
Z
x+1
x
1
dx
=
dx
+
dx
4 + x2
4 + x2
4 + x2
On the first integral on the right hand side, we use direct substitution with u = 4+x2 ,
and du = 2x dx. We get:
Z
Z
1
x
dx =
du = ln |2u| + C = ln(8 + 2x2 ) + C
2
4+x
2u
On the second integral on the right hand side, we use
inverse trigonometric substitux
), so 2 sec2 t dt = dx. Thus,
tion with 2 tan t = x (or equivalently, t = arctan
2
Z
Z
Z
1
1
2 sec2 t
2
dx
=
2
sec
t
dt
=
dt
4 + x2
4 + 4 tan2 t
4 sec2 t
Z
x
1
t
1
=
dt = + C = arctan
+C
2
2
2
2
Therefore,
Z
Z
Z
x+1
x
1
dx
=
dx
+
dx
4 + x2
4 + x2
4 + x2
Z
8)
= ln(8 + 2x2 ) +
x
1
arctan
+C
2
2
sin(tan θ)
dθ
cos2 θ
Solution: Using direct substitution with u = tan θ and du = sec2 θ dθ, we get:
Z
Z
Z
sin(tan θ)
2
dθ = sec θ sin(tan θ) dθ = sin u du = − cos u + C
cos2 θ
Z
sin(tan θ)
⇒
dθ = − cos(tan θ) + C
cos2 θ
Page 4 of 22
MATH 105 921
Solutions to Integration Exercises
√
x 3 − 2x − x2 dx
Z
9)
Solution: Completing the square, we get 3 − 2x − x2 = 4 − (x + 1)2 . Using direct
substitution with u = x + 1 and du = dx, we get:
Z √
Z
Z √
Z √
√
2
2
2
x 3 − 2x − x dx = (u − 1) 4 − u du = u 4 − u du −
4 − u2 du
For the first integral on the right hand side, using direct substitution with t = 4 − u2 ,
and dt = −2u du, we get:
Z √
Z
3
1√
1 3
1
2
u 4 − u du = −
t dt = − t 2 + C = − (4 − u2 ) 2 + C
2
3
3
For the second integral on the right hand
side, using inverse trigonometric substitution
u
with 2 sin s = u, that is, s = arcsin 2 , and 2 cos s ds = du, we get:
Z p
Z
Z √
2
2
4 − u du =
4 − 4 sin s2 cos s ds = 4 cos2 s ds
Z
1 + cos(2s)
= (2 + 2 cos(2s)) ds
(using half-angle formula cos2 s =
)
2
= 2s + sin(2s) + C
= 2s + 2 sin s cos s + C
(using double-angle formula sin(2s) = 2 sin s cos s)
u
u
u
+ 2 sin(arcsin
) cos(arcsin
)+C
= 2 arcsin
2
2
2
√
u
4 − u2
= 2 arcsin
+u
+C
2
2
Therefore,
Z √
Z √
Z √
x 3 − 2x − x2 dx = u 4 − u2 du −
4 − u2 du
√
u
3
1
4 − u2
2 2
= − (4 − u ) − 2 arcsin
−u
+C
3
2
2
!
p
Z √
2
4
−
(x
+
1)
3
1
x
+
1
− (x + 1)
+C
⇒ x 3 − 2x − x2 dx = − (4 − (x + 1)2 ) 2 − 2 arcsin
3
2
2
Z
10)
π
3
sin3 z cos z dz
0
Page 5 of 22
MATH 105 921
Solutions to Integration Exercises
Solution: Using direct substitution
√ with u = sin z, and du = cos z dz, when z = 0,
π
then u = 0, and when z = 3 , u = 23 . We have that:
Z
π
3
√
Z
3
sin z cos z dz =
0
Z
⇒
0
π
3
sin3 z cos z dz =
0
Z
11)
3x2
3
2
√
9
u4 23
|0 =
u du =
4
64
3
9
64
1
dx
+ 2x + 1
2
2
1
+ =
Solution: Completing the square, we get that 3x + 2x + 1 = 3 x +
3
3
!
2
1
3
2 9
1
x+
+ 1 . Using direct substitution with u = √ x +
, and du =
3 2
3
3
2
3
√ dx, we get:
2
Z
Z
Z
1
1
3
1
√
dx =
du = √ arctan u + C
dx =
9
1 2
2
3x + 2x + 1
2( 2 (x + 3 ) + 1)
2(u2 + 1)
2
Z
1
1
3
1
⇒
dx = √ arctan √ x +
+C
2
3x + 2x + 1
3
2
2
2
Z
12)
et
1
dt
+1
1
Solution: Using direct substitution with u = et + 1 and du = et dt, so dt = t du =
e
1
du. Hence, we get:
u−1
Z
Z
1
1
dt =
du
t
e +1
u(u − 1)
Using partial fraction, we get:
1
A
B
A(u − 1) + Bu
(A + B)s + (−A)
= +
=
=
u(u − 1)
u u−1
u(u − 1)
u(u − 1)
Page 6 of 22
MATH 105 921
Solutions to Integration Exercises
Thus, A + B = 0 and −A = 1. So, A = −1, and B = 1. Thus, we have that:
Z
Z
Z
−1
1
1
du =
du +
du
u(u − 1)
u
u−1
Therefore,
Z
Z
Z
1
−1
1
du =
du +
du = − ln |u| + ln |u − 1| + C
u(u − 1)
u
u−1
Z
1
⇒
dt = − ln |et + 1| + ln |et | + C = − ln |et + 1| + t + C
t
e +1
Z
13)
e3a cos(3a) da
Solution: Using direct substitution with t = 3a, and dt = 3 da, we get:
Z
Z
1 t
3a
e cos t dt
e cos(3a) da =
3
Using integration by parts with u = cos t, du = − sin t dt, and dv = et dt, v = et , we
get:
Z
Z
1 t
1
1 t
e cos t dt = e cos t +
et sin t dt
3
3
3
Using integration by parts again on the remaining integral with u1 = sin t, du1 =
cos t dt, and dv1 = et dt, v1 = et , we get:
Z
Z
1
1
1
t
t
e sin t dt = sin te −
et cos t dt
3
3
3
Thus,
Z
1 t
e cos t dt =
3
Z
1 t
⇒
e cos t dt =
3
1 t
e cos t +
3
1 t
e cos t +
6
1
1
sin tet −
3
3
1 t
e sin t + C
6
Z
et cos t dt
Therefore,
Z
1
1
e3a cos(3a) da = e3a cos(3a) + e3a sin(3a) + C
6
6
Page 7 of 22
MATH 105 921
Z
14)
Solutions to Integration Exercises
x2
dx
1 + x6
Solution: Using direct substitution with u = x3 , and du = 3x2 dx, we get:
Z
Z
x2
1
1
1
du
=
arctan
u
+
C
=
arctan(x3 ) + C
dx
=
1 + x6
3(1 + u2 )
3
3
Z
15)
1
dt
t(ln t)2
Solution: Using direct substitution with u = ln(t) and du = 1t dt, we get:
Z
Z
1
1
1
dt
=
du
=
−
+C
t(ln t)2
u2
u
Z
1
1
⇒
dt
=
−
+C
t(ln t)2
ln t
Z
16)
xe2x
dx
(2x + 1)2
Solution: Using integration by parts with u = xe2x , du = (e2x + 2xe2x ) dx, and
1
dv = (2x + 1)−2 dx, v = − 2(2x+1)
, we get:
Z
xe2x
xe2x
dx
=
−
+
(2x + 1)2
2(2x + 1)
Z
e2x + 2xe2x
dx
2(2x + 1)
On the remaining integral, using direct substitution with u = 2x + 1, and du = 2 dx,
we get:
Z 2x
Z u−1
Z
e + 2xe2x
e
+ (u − 1)eu−1
1 u−1
1
1
dx =
du =
e
du = eu−1 + C = e2x + C
2(2x + 1)
4u
4
4
4
Therefore,
Z
xe2x
xe2x
1 2x
e2x
dx
=
−
+
e
+
C
=
+C
(2x + 1)2
2(2x + 1) 4
4(2x + 1)
Page 8 of 22
MATH 105 921
Z
17)
Solutions to Integration Exercises
(tan x + cot x)2 dx
Solution:
Z
Z
2
(tan x + cot x) dx = (tan2 x + 2 tan x cot x + cot2 x) dx
Z
= (sec2 x − 1 + 2 + csc2 x − 1) dx
(using identities for tan2 x and cot2 x)
Z
= (sec2 x + csc2 x) dx
= tan x − cot x + C
Z
18)
2
tet sin(t2 ) dt
Solution: Using direct substitution with x = t2 and dx = 2t dt, we get:
Z
Z
1
t2
2
te sin(t ) dt =
ex sin x dx
2
Using integration by parts with u = sin x, du = cos x dx, and dv = ex dx, v = ex , we
get:
Z
Z
1 x
1
1 x
e sin x dx = e sin x −
ex cos x dx
2
2
2
Using integration by parts again on the remaining integral with u1 = cos x, du1 =
− sin x dx, and dv1 = ex dx, v1 = ex , we get:
Z
Z
1
1 x
1
x
e cos x dx = e cos x +
ex sin x dx
2
2
2
Thus,
Z
1 x
e sin x dx =
2
Z
1 x
⇒
e sin x dx =
2
1 x
e sin x −
2
1 x
e sin x −
4
1 x
1
e cos x −
2
2
1 x
e cos x + C
4
Z
ex sin x dx
Therefore,
Z
1 2
1 2
2
tet sin(t2 ) dt = et sin(t2 ) − et cos(t2 ) + C
4
4
Page 9 of 22
MATH 105 921
Solutions to Integration Exercises
2p − 4
dp
p2 − p
Z
19)
Solution: Using partial fraction, we get:
2p − 4
A
B
A(p − 1) + Bp
(A + B)p + (−A)
= +
=
=
p(p − 1)
p
p−1
p(p − 1)
p(p − 1)
Thus, A + B = 2 and −A = −4. So, A = 4, and B = −2. We have that:
Z
Z
Z
2p − 4
4
2
dp =
dp −
dp
p(p − 1)
p
p−1
Z
2p − 4
⇒
dp = 4 ln |p| − 2 ln |p − 1| + C
p(p − 1)
4
Z
20)
3
1
dx
(3x − 7)2
Solution: Using direct substitution with u = 3x − 7, and du = 3 dx, when x = 3,
then u = 2, and when x = 4, u = 5. We have that:
Z 4
Z 5
1
1
−1 5
1
1
1
dx =
du =
|2 = − + =
2
2
3u
15 6
10
3 (3x − 7)
2 3u
Z 4
1
1
dx
=
⇒
2
10
3 (3x − 7)
Z
21)
t3
5
(2 − t2 ) 2
dt
Solution: Using direct substitution with u = 2 − t2 , and du = −2t dt, we get:
Z
Z
Z
t3
t2
2−u
−
5 dt =
5 (t dt) =
5 du
(2 − t2 ) 2
(2 − t2 ) 2
2u 2
Z
5
1 3
= (−u− 2 + u− 2 ) du
2
1
2 3
= u− 2 − u− 2 + C
3
Z
3
1
t3
2
⇒
(2 − t2 )− 2 − (2 − t2 )− 2 + C
5 dt =
3
(2 − t2 ) 2
Page 10 of 22
MATH 105 921
Z
22)
x2
√
Solutions to Integration Exercises
1
dx
4 − x2
Solution:
Using inverse trigonometric substitution with x = 2 sin y, that is, y =
x
arcsin 2 , and dx = 2 cos y dy, we get:
Z
Z
Z
1
2 cos y
2 cos y
√
p
dx =
dy
dy
=
4 sin2 y(2 cos y)
x2 4 − x2
4 sin2 y 4 − 4 sin2 y
Z
1
1
csc2 y dy = − cot y + C
=
4
4
Therefore,
Z
√
x
1
1
4 − x2
√
)+C =−
+C
dx = − cot(arcsin
4
2
4x
x2 4 − x2
Z p
23)
y 2 − 1 dy
Solution:
Using inverse trigonometric substitution with y = sec u, that is, u =
arccos y1 , and dy = sec u tan u du, we get:
Z p
Z √
Z
2
2
y − 1 dy =
sec u − 1(sec u tan u du) = tan2 u sec u du
Z
Z
Z
2
3
= (sec u − 1) sec u du = sec u du − sec u du
For the second integral on the right hand side, we have that:
Z
sec u du = ln | sec u + tan u| + C
For the first integral on the right hand side, we use the reduction formula:
Z
Z
1
1
1
1
3
sec u du = tan u sec u + ln | sec u + tan u| + C
sec u du = tan u sec u +
2
2
2
2
p
Observe that since u = arccos y1 , we have that tan u = y 2 − 1. Therefore,
Z
Z
1
1
sec u du − sec u du = tan u sec u − ln | sec u + tan u| + C
2
2
Z p
p
p
1
1
⇒
y 2 − 1 dy = y y 2 − 1 − ln |y + y 2 − 1| + C
2
2
3
Page 11 of 22
MATH 105 921
Solutions to Integration Exercises
Z
24)
x sin x cos x dx
Solution: Using the double angle identity sin(2x) = 2 sin x cos x, we have that:
Z
Z
1
x sin(2x) dx
x sin x cos x dx =
2
Using direct substitution with t = 2x, and dt = 2 dx, we get:
Z
Z
1
1
x sin(2x) dx =
t sin t dt
2
8
Using integration by parts with u = t, du = dt, and dv = sin t dt, v = − cos t, we get:
Z
Z
1
1
1
1
1
t sin t dt = − t cos t +
cos t dt = − t cos t + sin t + C
8
8
8
8
8
Therefore,
Z
Z
25)
1
1
x sin x cos x dx = − x cos(2x) + sin(2x) + C
4
8
(1 + cos θ)2 dθ
Solution:
Z
Z
2
(1 + cos θ) dθ = (1 + 2 cos θ + cos2 θ) dθ
Z
Z
Z
= dθ + 2 cos θ dθ + cos2 θ dθ
Z 1 + cos(2θ)
dθ
= θ + 2 sin θ +
2
θ sin(2θ)
= θ + 2 sin θ + +
+C
2
4
Z
3
1
⇒ (1 + cos θ)2 dθ = θ + 2 sin θ + sin(2θ) + C
2
4
Z
26)
1
√
dx
4x − x2
Page 12 of 22
(using half-angle formula)
MATH 105 921
Solutions to Integration Exercises
Solution: Completing the square yields 4x − x2 = 4 − (x − 2)2 . Using direct substitution with u = x − 2, and du = dx, we get:
Z
Z
1
1
√
√
dx =
du
4x − x2
4 − u2
u
, and
Using inverse trigonometric substitution with u = 2 sin t, that is, t = arcsin
2
du = 2 cos t dt, we get:
Z
Z
Z
Z
2 cos t
2 cos t
1
√
p
dt =
dt = dt = t + C
du =
2 cos t
4 − u2
4 − sin2 t
Z
1
x−2
√
+C
dx = arcsin
⇒
2
4x − x2
1
Z
27)
1
1
0
1 + x3
dx
1 −2
x 3 dx, so dx =
3
2
3x 3 du = 3(u − 1)2 du. When x = 0, u = 1 and when x = 1, u = 2. We have that:
Z 1
Z 2
Z 2
1
3(u − 1)2
3
(3u − 6 + ) du
du =
1 dx =
u
u
0 1 + x3
1
1
3
= ( u2 − 6u + 3 ln |u|) |21
2
3
3
= (6 − 12 + 3 ln 2) − ( − 6 + 3 ln 1) = − + 3 ln 2
2
2
Z 1
1
3
⇒
+ 3 ln 2.
1 dx = −
2
0 1 + x3
1
Solution: Using direct substitution with u = 1 + x 3 , and du =
Z
28)
1
dx
x3 + x
Solution: Using partial fractions, we have:
1
A Bx + C
A(x2 + 1) + (Bx + C)x
(A + B)x2 + Cx + A
=
+
=
=
x3 + x
x
x2 + 1
x3 + x
x3 + x
So, A + B = 0, C = 0 and A = 1. So, B = −1 and we get:
Z
Z
Z
Z
1
1
x
x
dx =
dx −
dx = ln |x| −
dx
3
2
2
x +x
x
x +1
x +1
Page 13 of 22
MATH 105 921
Solutions to Integration Exercises
On the remaining integral, using direct substitution with u = x2 + 1 and du = 2x dx,
we get:
Z
Z
x
1
1
1
dx =
du = ln |u| + C = ln(x2 + 1) + C
2
x +1
2u
2
2
Therefore,
Z
x3
1
1
dx = ln |x| − ln(x2 + 1) + C
+x
2
Remark: This involves partial fractions with non-linear factors, which you are not
required to master in this course!
Z
29)
ln(1 + t) dt
Solution: Using direct substitution with s = 1 + t, and ds = dt, we have that:
Z
Z
ln(1 + t) dt = ln s ds
1
ds, and dv = ds, v = s, we get:
s
Z
Z
Z
1
ln s ds = s ln s − s ds = s ln s −
ds = s ln s − s + C
s
Using integration by parts with u = ln s, du =
Therefore,
Z
ln(1 + t) dt = (1 + t) ln(1 + t) − (1 + t) + C
Z
30)
sin(3x) cos(5x) dx
Solution: Using the trigonometric identity that sin a cos b = 12 (sin(a+b)+sin(a−b)),
we get:
Z
Z
1
1
1
sin(3x) cos(5x) dx =
(sin(8x) + sin(−2x)) dx = − cos(8x) + cos(−2x) + C
2
16
4
Remark: You are not required to memorize any sum to product or product to sum
trigonometric identities!
Page 14 of 22
MATH 105 921
Z
31)
k2
Solutions to Integration Exercises
1
dk
− 6k + 9
Solution: By completing the square, we observe that k 2 − 6k + 9 = (k − 3)2 . So,
using direct substitution with u = k − 3, and du = dk, we have that:
Z
Z
Z
1
1
1
1
dk
=
+C
dk
=
du
=
−
k 2 − 6k + 9
(k − 3)2
u2
u
Z
1
1
⇒
dk = −
+C
2
k − 6k + 9
k−3
Z
32)
1
dx
sec x − 1
1
, we get:
cos x
Z
Z
Z Z
1
cos x
1
1
dx =
dx =
dx
−1 +
dx = −x +
sec x − 1
1 − cos x
1 − cos x
1 − cos x
x
For the remaining integral, use a direct substitution with t = tan
, so dt =
2
√
1
1
x
x
x
sec2
dx. We also can compute that sec
= t2 + 1, cos
= √
2
2
2
2
t2 + 1
x
t
2
dt. Using double angle formula, we get:
and sin
=√
. So, dx = 2
2
t +1
t2 + 1
Solution: Since sec x =
2
cos x = cos
x
2
2
− sin
x
2
=
1
t2
1 − t2
−
=
t2 + 1 t2 + 1
t2 + 1
So, after the substitution, we get:
Z
Z
Z
2
1
1
1
dx =
dt =
dt
2
2
1−t
1 − cos x
t2
1 − t2 +1 t + 1
x
1
= − + C = − cot
+C
t
2
Therefore,
Z
x
1
dx = −x − cot
+C
sec x − 1
2
Remark: This is an extremely challenging question; do not panic if you do not know
how to solve it!
Page 15 of 22
MATH 105 921
1
Z
Solutions to Integration Exercises
2
dx
+1
33)
e−x
0
Solution: Using direct substitution with u = e−x + 1, and du = −e−x dx, that is
1
dx = − u−1
du. When x = 0, u = 2, and when x = 1, u = e−1 + 1. So, we get:
Z
0
1
2
dx =
e−x + 1
Z
e−1 +1
2
−2
du
u(u − 1)
Using partial fraction, we get:
−2
A
B
A(u − 1) + Bu
(A + B)s + (−A)
= +
=
=
u(u − 1)
u u−1
u(u − 1)
u(u − 1)
Thus, A + B = 0 and −A = −2. So, A = 2, and B = −2. Thus, we have that:
Z
2
e−1 +1
−2
du =
u(u − 1)
Z
2
e−1 +1
2
du −
u
Z
2
e−1 +1
2
du
u−1
Therefore,
Z
e−1 +1
2
Z
34)
c2
−2
−1
du = (2 ln |u| − 2 ln |u − 1|) |e2 +1 = (2 ln(e−1 + 1) + 2) − (2 ln 2 − 0)
u(u − 1)
Z
2
⇒
dx = 2 ln(e−1 + 1) + 2 − 2 ln 2.
−x
e +1
1
dc
− 6c + 10
Solution: Completing the square yields c2 − 6c + 10 = (c − 3)2 + 1. So, using direct
substitution with u = c − 3, and du = dc, we have that:
Z
Z
Z
1
1
1
dc =
dc =
du = arctan u + C
2
2
2
c − 6c + 10
(c − 3) + 1
u +1
Z
1
⇒
dc = arctan(c − 3) + C
2
c − 6c + 10
Z
35)
f (x)f 0 (x) dx
Page 16 of 22
MATH 105 921
Solutions to Integration Exercises
Solution: Using direct substitution with u = f (x), and du = f 0 (x) dx, we get:
Z
Z
1
0
f (x)f (x) dx = u du = u2 + C
2
Z
1
⇒ f (x)f 0 (x) dx = (f (x))2 + C
2
Z
36)
x2
1
dx
+ 4x + 5
Solution: Completing the square, we get x2 + 4x + 5 = (x + 2)2 + 1. Using direct
substitution with u = x + 2 and du = dx, we get:
Z
Z
Z
1
1
1
dx =
dx =
du = arctan(u) + C
2
2
2
x + 4x + 5
(x + 2) + 1
u +1
Z
1
⇒
dx = arctan(x + 2) + C
x2 + 4x + 5
2
Z
37)
0
1
dx
(3 + 5x)2
Solution: Using direct substitution with u = 3 + 5x, and du = 5 dx, when x = 0,
then u = 3, and when x = 2, u = 13. We have that:
Z 13
Z 2
1
1
−1 13
1
1
2
dx =
du =
|3 = − +
=
2
2
5u
5u
65 15
39
3
0 (3 + 5x)
Z 2
1
2
⇒
dx =
2
39
0 (3 + 5x)
Z
38)
sin(ln u) du
Solution: Using direct substitution with t = ln u, that is, u = et , and du = et dt, we
have that:
Z
Z
sin(ln u) du = et sin t dt
Page 17 of 22
MATH 105 921
Solutions to Integration Exercises
Using integration by parts twice to compute the integral on the right hand side (see
the solution of question 18 for details), we have that:
Z
1
1
et sin t dt = et sin t − et cos t + C
2
2
Therefore,
Z
1
1
1
1
sin(ln u) du = eln u sin(ln u) − eln u cos(ln u) + C = u sin(ln u) − u cos(ln u) + C
2
2
2
2
Z
39)
r(ln r)2 dr
Solution: Using integration by parts with u = (ln r)2 , du =
v=
r2
, we get that:
2
Z
r2 (ln r)2
r(ln r) dr =
−
2
2
2 ln r
dr, and dv = r dr,
r
Z
r ln r dr
Using integration by parts again on the remaining integral with u1 = ln r, du1 =
and dv1 = r dr, v1 =
Z
r2
, we get that:
2
r2 ln r
r ln r dr =
−
2
Z
r
r2 ln r r2
dr =
−
+C
2
2
4
Therefore,
Z
Z
40)
r2 (ln r)2 r2 ln r r2
r(ln r) dr =
−
+
+C
2
2
4
2
1
dx
x3 − x
Page 18 of 22
1
dr,
r
MATH 105 921
Solutions to Integration Exercises
Solution: Using partial fraction, we get:
1
A
B
C
A(x2 − 1) + B(x2 − x) + C(x2 + x)
=
+
+
=
x3 − x
x x+1 x−1
x3 − x
2
(A + B + C)x + (C − B)x + (−A)
=
x3 − x
Thus, A + B + C = 0, C − B = 0 and −A = 1. Therefore, A = −1, and B + C = 1,
which gives C = 12 and B = − 12 . So,
Z
Z
Z
Z
1
1
1
1
dx = − dx −
dx +
dx
3
x −x
x
2(x + 1)
2(x − 1)
Z
1
1
1
⇒
dx = − ln |x| − ln |x + 1| + ln |x − 1| + C
3
x −x
2
2
Remark: This involves partial fractions with 3 distinct roots in the denominator, which
you are not required to master in this course!
Z
41)
sec3 u du
Solution: We use the reduction formula:
Z
Z
1
1
1
1
3
sec u du = tan u sec u + ln | sec u + tan u| + C
sec u du = tan u sec u +
2
2
2
2
Z √
42)
x2 − 2x − 8
dx
x−1
Solution: Observe that x2 − 2x − 8 = (x − 1)2 − 9. Using direct substitution with
t = x − 1, and dt = dx, we get:
Z √ 2
Z √2
x − 2x − 8
t −9
dx =
dt
x−1
t
Using inverse trigonometric substitution with t = 3 sec y, and dt = 3 sec y tan y dy, we
Page 19 of 22
MATH 105 921
Solutions to Integration Exercises
get:
Z √
t2 − 9
dt =
t
Z p
9 sec2 y − 9
3 sec y tan y dy =
3 sec y
Z
3 tan2 y dy
Z
3(sec2 y − 1) dy = 3 tan y − 3y + C
Z √ 2
√
x − 2x − 8
3
3
3
2
dx = 3 tan(arccos
) − 3 arccos
+ C = t − 9 − 3 arccos
+C
⇒
x−1
t
t
t
p
3
2
= (x − 1) − 9 − 3 arccos
+C
x−1
=
Z √
43)
r2 − 1
dr
r
Solution:
Using inverse trigonometric substitution with sec s = r, that is, s =
1
arccos
, and sec s tan s ds = dr, we get:
r
Z √ 2
Z √ 2
Z
r −1
sec s − 1
dr =
sec tan s ds = tan2 s ds
r
sec s
Z
= (sec2 s − 1) ds = tan s − s + C
Z √ 2
√
r −1
1
1
1
⇒
dr = tan(arccos
) − arccos
+ C = r2 − 1 − arccos
+C
r
r
r
r
Z
44)
2
2
(et + 16)tet dt
2
2
Solution: Using direct substitution with u = et and du = 2tet dt, we get:
Z
Z
1
1
t2
t2
(e + 16)te dt =
(u + 16) du = u2 + 8u + C
2
4
Z
1 2
2
2
2
⇒ (et + 16)tet dt = e2t + 8et + C
4
Z
45)
√
y ln y dy
Page 20 of 22
MATH 105 921
Solutions to Integration Exercises
Solution: Using integration by parts with u = ln y, du =
2 3
v = y 2 , we get:
3
Z
√
Z
46)
2 3
y ln y dy = y 2 ln y −
3
Z
1
√
dy and dv = y dy,
y
2 1
2 3
4 3
y 2 dy = y 2 ln y − y 2 + C
3
3
9
cos θ
dθ
1 + sin2 θ
Solution: Using direct substitution with u = sin θ, and du = cos θ dθ, we have that:
Z
Z
cos θ
1
du = arctan u + C
2 dθ =
1 + u2
1
+
sin
θ
Z
cos θ
dθ = arctan(sin θ) + C
⇒
1 + sin2 θ
Z
47)
x2
√
1
dx
x2 + 4
Solution: Using inverse trigonometric substitution with 2 tan u = x, and 2 sec u du =
dx, we get:
Z
Z
Z
1
1
2 sec2 u
2
√
√
dx =
2
sec
u
du
=
du
8 tan2 u sec u
x2 x2 + 4
4 tan2 u 4 tan2 u + 4
Z
Z
cos2 u
1
=
du =
cot u csc u du
2
4
4 cos u sin u
x
1
1
= − csc u + C = − csc(arctan
)+C
4
4
2
√
Z
1
x2 + 4
√
⇒
+C
dx = −
4x
x2 x2 + 4
Z
48)
2
tet dt
Page 21 of 22
MATH 105 921
Solutions to Integration Exercises
Solution: Using direct substitution with u = t2 , and du = 2t dt, we have that:
Z
Z
1 u
1
t2
te dt =
e du = eu + C
2
2
Z
1
2
2
⇒ tet dt = et + C
2
Z
49)
cos(πt) cos(sin(πt)) dt
Solution: Using direct substitution with u = sin(πt) dt, and du = π cos(πt) dt, we
have that:
Z
Z
1
1
cos(πt) cos(sin(πt)) dt =
cos u du = sin u + C
π
π
Z
1
⇒ cos(πt) cos(sin(πt)) dt = sin(cos(πt)) + C
π
π
4
Z
50)
sin5 (x) dx
0
Solution: Using direct substitution with u = cos x, and du = − sin x dx, when x = 0,
1
π
then u = 1, and when x = , u = √ . We have that:
4
2
Z π
Z π
Z π
4
4
4
5
2
2
sin (x) dx =
(sin (x)) sin x dx =
(1 − cos2 x)2 sin x dx
0
0
Z
=
1
Z
⇒
0
π
4
0
√1
2
−(1 − u2 )2 du =
Z
√1
2
(−1 + 2u2 − u4 ) du
1
√1
2
1
= (−u + u3 − u5 ) |1 2
3
5
1
1
2 1
8
1
43
= (− √ + √ − √ ) − (−1 + − ) = − √ +
3 5
2 3 2 20 2
60 2 15
8
43
sin5 (x) dx = − √ +
60 2 15
Page 22 of 22