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MATH 105 921 Solutions to Integration Exercises
MATH 105 921 Solutions to Integration Exercises
1)
∫s2 + 1
s2 − 1ds
Solution: Performing polynomial long division, we have that:∫s2 + 1
s2 − 1ds =
∫(1 +
2
s2 − 1) ds
=
∫ds+
∫2
s2 − 1ds
= s+
∫2
s2 − 1ds
Using partial fraction on the remaining integral, we get:
2
s2 − 1=
A
s− 1+
B
s+ 1=A(s+ 1) +B(s− 1)
(s+ 1)(s− 1)=
(A+B)s+ (A−B)
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:∫
2
s2 − 1ds =
∫1
s− 1ds−
∫1
s+ 1ds
Therefore, ∫s2 + 1
s2 − 1ds = s+
∫2
s2 − 1ds
= s+
∫1
s− 1ds−
∫1
s+ 1ds
= s+ ln |s− 1| − ln |s+ 1|+ C
2)
∫ 0
4
x√
1 + 2x dx
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,∫ 0
4
x√
1 + 2x dx =
∫ 1
9
1
4(u− 1)
√u du
=
∫ 1
9
1
4(u
32 − u
12 ) du
= (1
10u
52 − 1
6u
32 ) |19
= (1
10− 1
6)− (
243
10− 27
6)
=−298
15
3)
∫sin2 x cos2 x dx
Solution: Using half-angle identities sin2 x = 1−cos(2x)2
and cos2 x = 1+cos(2x)2
, we get:∫sin2 x cos2 x dx =
∫1
4(1− cos(2x))(1 + cos(2x)) dx
=
∫1
4(1− cos2(2x)) dx
=
∫1
4dx−
∫1
4cos2(2x) dx
=x
4− 1
4
∫cos2(2x) dx
On the remaining integral, we apply the half-angle identity cos2(2x) = 1+cos(4x)2
, andobtain: ∫
cos2(2x) dx =
∫1 + cos(4x)
2dx =
x
2+
1
8sin(4x) + C
Hence, ∫sin2 x cos2 x dx =
x
4− 1
4(x
2+
1
8sin(4x)) + C =
x
8− 1
32sin(4x) + C
4)
∫sin(√w) dw
Page 2 of 22
MATH 105 921 Solutions to Integration Exercises
Solution: Using direct substitution with t =√w, and dt = 1
2√wdw, that is, dw =
2√w dt = 2t dt, we get: ∫
sin(√w) dw =
∫2t sin t dt
Using integration by part method with u = 2t and dv = sin t dt, so du = 2 dt andv = − cos t, we get:∫
2t sin t dt = −2t cos t+
∫2 cos t dt = −2t cos t+ 2 sin t+ C
Therefore, ∫sin(√w) dw = −2
√w cos(
√w) + 2 sin(
√w) + C
5)
∫ln(x)
xdx
Solution: Using direct substitution with u = ln(x) and du = 1xdx, we get:∫
ln(x)
xdx =
∫u du =
u2
2+ C
⇒∫
ln(x)
xdx =
1
2(ln(x))2 + C
6)
∫sin t cos(2t) dt
Solution: Recall the double-angle formula that cos(2t) = 2 cos2 t− 1, we get:∫sin t cos(2t) dt =
∫sin t(2 cos2 t− 1) dt
=
∫2 sin t cos2 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:∫
2 sin t cos2 t dt =
∫−2u2 du = −2
3u3 + C = −2
3cos3 t+ C
Page 3 of 22
MATH 105 921 Solutions to Integration Exercises
Therefore, ∫sin t cos(2t) dt = −2
3cos3 t+ cos t+ C
7)
∫x+ 1
4 + x2dx
Solution: Observe that we may split the integral as follows:∫x+ 1
4 + x2dx =
∫x
4 + x2dx+
∫1
4 + x2dx
On the first integral on the right hand side, we use direct substitution with u = 4+x2,and du = 2x dx. We get:∫
x
4 + x2dx =
∫1
2udu = ln |2u|+ C = ln(8 + 2x2) + C
On the second integral on the right hand side, we use inverse trigonometric substitu-
tion with 2 tan t = x (or equivalently, t = arctan(x
2
)), so 2 sec2 t dt = dx. Thus,
∫1
4 + x2dx =
∫1
4 + 4 tan2 t2 sec2 t dt =
∫2 sec2 t
4 sec2 tdt
=
∫1
2dt =
t
2+ C =
1
2arctan
(x2
)+ C
Therefore,∫x+ 1
4 + x2dx =
∫x
4 + x2dx+
∫1
4 + x2dx = ln(8 + 2x2) +
1
2arctan
(x2
)+ C
8)
∫sin(tan θ)
cos2 θdθ
Solution: Using direct substitution with u = tan θ and du = sec2 θ dθ, we get:∫sin(tan θ)
cos2 θdθ =
∫sec2 θ sin(tan θ) dθ =
∫sinu du = − cosu+ C
⇒∫
sin(tan θ)
cos2 θdθ = − cos(tan θ) + C
Page 4 of 22
MATH 105 921 Solutions to Integration Exercises
9)
∫x√
3− 2x− x2 dx
Solution: Completing the square, we get 3 − 2x − x2 = 4 − (x + 1)2. Using directsubstitution with u = x+ 1 and du = dx, we get:∫
x√
3− 2x− x2 dx =
∫(u− 1)
√4− u2 du =
∫u√
4− u2 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:∫
u√
4− u2 du =
∫−1
2
√t dt = −1
3t32 + C = −1
3(4− u2)
32 + C
For the second integral on the right hand side, using inverse trigonometric substitutionwith 2 sin s = u, that is, s = arcsin
(u2
), and 2 cos s ds = du, we get:∫ √
4− u2 du =
∫ √4− 4 sin2 s2 cos s ds =
∫4 cos2 s ds
=
∫(2 + 2 cos(2s)) ds (using half-angle formula cos2 s =
1 + cos(2s)
2)
= 2s+ sin(2s) + C
= 2s+ 2 sin s cos s+ C (using double-angle formula sin(2s) = 2 sin s cos s)
= 2 arcsin(u
2
)+ 2 sin(arcsin
(u2
)) cos(arcsin
(u2
)) + C
= 2 arcsin(u
2
)+ u
(√4− u2
2
)+ C
Therefore,∫x√
3− 2x− x2 dx =
∫u√
4− u2 du−∫ √
4− u2 du
= −1
3(4− u2)
32 − 2 arcsin
(u2
)− u
(√4− u2
2
)+ C
⇒∫x√
3− 2x− x2 dx = −1
3(4− (x+ 1)2)
32 − 2 arcsin
(x+ 1
2
)− (x+ 1)
(√4− (x+ 1)2
2
)+ C
10)
∫ π3
0
sin3 z cos z dz
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,then u = 0, and when z = π
3, u =
√32
. We have that:
∫ π3
0
sin3 z cos z dz =
∫ √3
2
0
u3 du =u4
4|√3
20 =
9
64
⇒∫ π
3
0
sin3 z cos z dz =9
64
11)
∫1
3x2 + 2x+ 1dx
Solution: Completing the square, we get that 3x2 + 2x + 1 = 3
(x+
1
3
)2
+2
3=
2
3
(9
2
(x+
1
3
)2
+ 1
). Using direct substitution with u =
3√2
(x+
1
3
), and du =
3√2dx, we get:
∫1
3x2 + 2x+ 1dx =
∫3
2(92(x+ 1
3)2 + 1)
dx =
∫1√
2(u2 + 1)du =
1√2
arctanu+ C
⇒∫
1
3x2 + 2x+ 1dx =
1√2
arctan
(3√2
(x+
1
3
))+ C
12)
∫1
et + 1dt
Solution: Using direct substitution with u = et + 1 and du = et dt, so dt =1
etdu =
1
u− 1du. Hence, we get: ∫
1
et + 1dt =
∫1
u(u− 1)du
Using partial fraction, we get:
1
u(u− 1)=A
u+
B
u− 1=A(u− 1) +Bu
u(u− 1)=
(A+B)s+ (−A)
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:∫1
u(u− 1)du =
∫−1
udu+
∫1
u− 1du
Therefore,∫1
u(u− 1)du =
∫−1
udu+
∫1
u− 1du = − ln |u|+ ln |u− 1|+ C
⇒∫
1
et + 1dt = − ln |et + 1|+ ln |et|+ C = − ln |et + 1|+ t+ C
13)
∫e3a cos(3a) da
Solution: Using direct substitution with t = 3a, and dt = 3 da, we get:∫e3a cos(3a) da =
∫1
3et cos t dt
Using integration by parts with u = cos t, du = − sin t dt, and dv = et dt, v = et, weget: ∫
1
3et cos t dt =
1
3et cos t+
1
3
∫et sin t dt
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:
1
3
∫et sin t dt =
1
3sin tet − 1
3
∫et cos t dt
Thus, ∫1
3et cos t dt =
1
3et cos t+
1
3sin tet − 1
3
∫et cos t dt
⇒∫
1
3et cos t dt =
1
6et cos t+
1
6et sin t+ C
Therefore, ∫e3a cos(3a) da =
1
6e3a cos(3a) +
1
6e3a sin(3a) + C
Page 7 of 22
MATH 105 921 Solutions to Integration Exercises
14)
∫x2
1 + x6dx
Solution: Using direct substitution with u = x3, and du = 3x2 dx, we get:∫x2
1 + x6dx =
∫1
3(1 + u2)du =
1
3arctanu+ C =
1
3arctan(x3) + C
15)
∫1
t(ln t)2dt
Solution: Using direct substitution with u = ln(t) and du = 1tdt, we get:∫
1
t(ln t)2dt =
∫1
u2du = −1
u+ C
⇒∫
1
t(ln t)2dt = − 1
ln t+ C
16)
∫xe2x
(2x+ 1)2dx
Solution: Using integration by parts with u = xe2x, du = (e2x + 2xe2x) dx, anddv = (2x+ 1)−2 dx, v = − 1
2(2x+1), we get:∫
xe2x
(2x+ 1)2dx = − xe2x
2(2x+ 1)+
∫e2x + 2xe2x
2(2x+ 1)dx
On the remaining integral, using direct substitution with u = 2x+ 1, and du = 2 dx,we get:∫e2x + 2xe2x
2(2x+ 1)dx =
∫eu−1 + (u− 1)eu−1
4udu =
∫1
4eu−1 du =
1
4eu−1 + C =
1
4e2x + C
Therefore, ∫xe2x
(2x+ 1)2dx = − xe2x
2(2x+ 1)+
1
4e2x + C =
e2x
4(2x+ 1)+ C
Page 8 of 22
MATH 105 921 Solutions to Integration Exercises
17)
∫(tanx+ cotx)2 dx
Solution:∫(tanx+ cotx)2 dx =
∫(tan2 x+ 2 tanx cotx+ cot2 x) dx
=
∫(sec2 x− 1 + 2 + csc2 x− 1) dx (using identities for tan2 x and cot2 x)
=
∫(sec2 x+ csc2 x) dx
= tanx− cotx+ C
18)
∫tet
2
sin(t2) dt
Solution: Using direct substitution with x = t2 and dx = 2t dt, we get:∫tet
2
sin(t2) dt =1
2
∫ex sinx dx
Using integration by parts with u = sinx, du = cosx dx, and dv = ex dx, v = ex, weget: ∫
1
2ex sinx dx =
1
2ex sinx− 1
2
∫ex cosx dx
Using integration by parts again on the remaining integral with u1 = cos x, du1 =− sinx dx, and dv1 = ex dx, v1 = ex, we get:
1
2
∫ex cosx dx =
1
2ex cosx+
1
2
∫ex sinx dx
Thus, ∫1
2ex sinx dx =
1
2ex sinx− 1
2ex cosx− 1
2
∫ex sinx dx
⇒∫
1
2ex sinx dx =
1
4ex sinx− 1
4ex cosx+ C
Therefore, ∫tet
2
sin(t2) dt =1
4et
2
sin(t2)− 1
4et
2
cos(t2) + C
Page 9 of 22
MATH 105 921 Solutions to Integration Exercises
19)
∫2p− 4
p2 − pdp
Solution: Using partial fraction, we get:
2p− 4
p(p− 1)=A
p+
B
p− 1=A(p− 1) +Bp
p(p− 1)=
(A+B)p+ (−A)
p(p− 1)
Thus, A+B = 2 and −A = −4. So, A = 4, and B = −2. We have that:∫2p− 4
p(p− 1)dp =
∫4
pdp−
∫2
p− 1dp
⇒∫
2p− 4
p(p− 1)dp = 4 ln |p| − 2 ln |p− 1|+ C
20)
∫ 4
3
1
(3x− 7)2dx
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:∫ 4
3
1
(3x− 7)2dx =
∫ 5
2
1
3u2du =
−1
3u|52= −
1
15+
1
6=
1
10
⇒∫ 4
3
1
(3x− 7)2dx =
1
10
21)
∫t3
(2− t2) 52
dt
Solution: Using direct substitution with u = 2− t2, and du = −2t dt, we get:∫t3
(2− t2) 52
dt =
∫t2
(2− t2) 52
(t dt) =
∫−2− u
2u52
du
=
∫(−u−
52 +
1
2u−
32 ) du
=2
3u−
32 − u−
12 + C
⇒∫
t3
(2− t2) 52
dt =2
3(2− t2)−
32 − (2− t2)−
12 + C
Page 10 of 22
MATH 105 921 Solutions to Integration Exercises
22)
∫1
x2√
4− x2dx
Solution: Using inverse trigonometric substitution with x = 2 sin y, that is, y =arcsin
(x2
), and dx = 2 cos y dy, we get:∫
1
x2√
4− x2dx =
∫2 cos y
4 sin2 y√
4− 4 sin2 ydy =
∫2 cos y
4 sin2 y(2 cos y)dy
=
∫1
4csc2 y dy = −1
4cot y + C
Therefore, ∫1
x2√
4− x2dx = −1
4cot(arcsin
(x2
)) + C = −
√4− x24x
+ C
23)
∫ √y2 − 1 dy
Solution: Using inverse trigonometric substitution with y = secu, that is, u =
arccos(
1y
), and dy = secu tanu du, we get:∫ √
y2 − 1 dy =
∫ √sec2 u− 1(secu tanu du) =
∫tan2 u secu du
=
∫(sec2 u− 1) secu du =
∫sec3 u du−
∫secu du
For the second integral on the right hand side, we have that:∫secu du = ln | secu+ tanu|+ C
For the first integral on the right hand side, we use the reduction formula:∫sec3 u du =
1
2tanu secu+
1
2
∫secu du =
1
2tanu secu+
1
2ln | secu+ tanu|+ C
Observe that since u = arccos(
1y
), we have that tanu =
√y2 − 1. Therefore,∫
sec3 u du−∫
secu du =1
2tanu secu− 1
2ln | secu+ tanu|+ C
⇒∫ √
y2 − 1 dy =1
2y√y2 − 1− 1
2ln |y +
√y2 − 1|+ C
Page 11 of 22
MATH 105 921 Solutions to Integration Exercises
24)
∫x sinx cosx dx
Solution: Using the double angle identity sin(2x) = 2 sinx cosx, we have that:∫x sinx cosx dx =
1
2
∫x sin(2x) dx
Using direct substitution with t = 2x, and dt = 2 dx, we get:
1
2
∫x sin(2x) dx =
1
8
∫t sin t dt
Using integration by parts with u = t, du = dt, and dv = sin t dt, v = − cos t, we get:
1
8
∫t sin t dt = −1
8t cos t+
1
8
∫cos t dt = −1
8t cos t+
1
8sin t+ C
Therefore, ∫x sinx cosx dx = −1
4x cos(2x) +
1
8sin(2x) + C
25)
∫(1 + cos θ)2 dθ
Solution:∫(1 + cos θ)2 dθ =
∫(1 + 2 cos θ + cos2 θ) dθ
=
∫dθ + 2
∫cos θ dθ +
∫cos2 θ dθ
= θ + 2 sin θ +
∫ (1 + cos(2θ)
2
)dθ (using half-angle formula)
= θ + 2 sin θ +θ
2+
sin(2θ)
4+ C
⇒∫
(1 + cos θ)2 dθ =3
2θ + 2 sin θ +
1
4sin(2θ) + C
26)
∫1√
4x− x2dx
Page 12 of 22
MATH 105 921 Solutions to Integration Exercises
Solution: Completing the square yields 4x− x2 = 4− (x− 2)2. Using direct substi-tution with u = x− 2, and du = dx, we get:∫
1√4x− x2
dx =
∫1√
4− u2du
Using inverse trigonometric substitution with u = 2 sin t, that is, t = arcsin(u
2
), and
du = 2 cos t dt, we get:∫1√
4− u2du =
∫2 cos t√4− sin2 t
dt =
∫2 cos t
2 cos tdt =
∫dt = t+ C
⇒∫
1√4x− x2
dx = arcsin
(x− 2
2
)+ C
27)
∫ 1
0
1
1 + x13
dx
Solution: Using direct substitution with u = 1 + x13 , and du =
1
3x−23 dx, so dx =
3x23 du = 3(u− 1)2 du. When x = 0, u = 1 and when x = 1, u = 2. We have that:∫ 1
0
1
1 + x13
dx =
∫ 2
1
3(u− 1)2
udu =
∫ 2
1
(3u− 6 +3
u) du
= (3
2u2 − 6u+ 3 ln |u|) |21
= (6− 12 + 3 ln 2)− (3
2− 6 + 3 ln 1) = −3
2+ 3 ln 2
⇒∫ 1
0
1
1 + x13
dx = −3
2+ 3 ln 2.
28)
∫1
x3 + xdx
Solution: Using partial fractions, we have:
1
x3 + x=A
x+Bx+ C
x2 + 1=A(x2 + 1) + (Bx+ C)x
x3 + x=
(A+B)x2 + Cx+ A
x3 + x
So, A+B = 0, C = 0 and A = 1. So, B = −1 and we get:∫1
x3 + xdx =
∫1
xdx−
∫x
x2 + 1dx = ln |x| −
∫x
x2 + 1dx
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: ∫
x
x2 + 1dx =
∫1
2udu =
1
2ln |u|+ C =
1
2ln(x2 + 1) + C
Therefore, ∫1
x3 + xdx = ln |x| − 1
2ln(x2 + 1) + C
Remark: This involves partial fractions with non-linear factors, which you are notrequired to master in this course!
29)
∫ln(1 + t) dt
Solution: Using direct substitution with s = 1 + t, and ds = dt, we have that:∫ln(1 + t) dt =
∫ln s ds
Using integration by parts with u = ln s, du =1
sds, and dv = ds, v = s, we get:∫
ln s ds = s ln s−∫s
1
sds = s ln s−
∫ds = s ln s− s+ C
Therefore, ∫ln(1 + t) dt = (1 + t) ln(1 + t)− (1 + t) + C
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:∫sin(3x) cos(5x) dx =
∫1
2(sin(8x) + sin(−2x)) dx = − 1
16cos(8x) +
1
4cos(−2x) + C
Remark: You are not required to memorize any sum to product or product to sumtrigonometric identities!
Page 14 of 22
MATH 105 921 Solutions to Integration Exercises
31)
∫1
k2 − 6k + 9dk
Solution: By completing the square, we observe that k2 − 6k + 9 = (k − 3)2. So,using direct substitution with u = k − 3, and du = dk, we have that:∫
1
k2 − 6k + 9dk =
∫1
(k − 3)2dk =
∫1
u2du = −1
u+ C
⇒∫
1
k2 − 6k + 9dk = − 1
k − 3+ C
32)
∫1
secx− 1dx
Solution: Since sec x =1
cosx, we get:∫
1
secx− 1dx =
∫cosx
1− cosxdx =
∫ (−1 +
1
1− cosx
)dx = −x+
∫1
1− cosxdx
For the remaining integral, use a direct substitution with t = tan(x
2
), so dt =
1
2sec2
(x2
)dx. We also can compute that sec
(x2
)=√t2 + 1, cos
(x2
)=
1√t2 + 1
and sin(x
2
)=
t√t2 + 1
. So, dx =2
t2 + 1dt. Using double angle formula, we get:
cosx = cos2(x
2
)− sin2
(x2
)=
1
t2 + 1− t2
t2 + 1=
1− t2
t2 + 1
So, after the substitution, we get:∫1
1− cosxdx =
∫1
1− 1−t2t2+1
(2
t2 + 1
)dt =
∫1
t2dt
= −1
t+ C = − cot
(x2
)+ C
Therefore, ∫1
secx− 1dx = −x− cot
(x2
)+ C
Remark: This is an extremely challenging question; do not panic if you do not knowhow to solve it!
Page 15 of 22
MATH 105 921 Solutions to Integration Exercises
33)
∫ 1
0
2
e−x + 1dx
Solution: Using direct substitution with u = e−x + 1, and du = −e−x dx, that isdx = − 1
u−1 du. When x = 0, u = 2, and when x = 1, u = e−1 + 1. So, we get:∫ 1
0
2
e−x + 1dx =
∫ e−1+1
2
−2
u(u− 1)du
Using partial fraction, we get:
−2
u(u− 1)=A
u+
B
u− 1=A(u− 1) +Bu
u(u− 1)=
(A+B)s+ (−A)
u(u− 1)
Thus, A+B = 0 and −A = −2. So, A = 2, and B = −2. Thus, we have that:∫ e−1+1
2
−2
u(u− 1)du =
∫ e−1+1
2
2
udu−
∫ e−1+1
2
2
u− 1du
Therefore,∫ e−1+1
2
−2
u(u− 1)du = (2 ln |u| − 2 ln |u− 1|) |e−1+1
2 = (2 ln(e−1 + 1) + 2)− (2 ln 2− 0)
⇒∫
2
e−x + 1dx = 2 ln(e−1 + 1) + 2− 2 ln 2.
34)
∫1
c2 − 6c+ 10dc
Solution: Completing the square yields c2 − 6c+ 10 = (c− 3)2 + 1. So, using directsubstitution with u = c− 3, and du = dc, we have that:∫
1
c2 − 6c+ 10dc =
∫1
(c− 3)2 + 1dc =
∫1
u2 + 1du = arctanu+ C
⇒∫
1
c2 − 6c+ 10dc = arctan(c− 3) + C
35)
∫f(x)f ′(x) dx
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MATH 105 921 Solutions to Integration Exercises
Solution: Using direct substitution with u = f(x), and du = f ′(x) dx, we get:∫f(x)f ′(x) dx =
∫u du =
1
2u2 + C
⇒∫f(x)f ′(x) dx =
1
2(f(x))2 + C
36)
∫1
x2 + 4x+ 5dx
Solution: Completing the square, we get x2 + 4x + 5 = (x + 2)2 + 1. Using directsubstitution with u = x+ 2 and du = dx, we get:∫
1
x2 + 4x+ 5dx =
∫1
(x+ 2)2 + 1dx =
∫1
u2 + 1du = arctan(u) + C
⇒∫
1
x2 + 4x+ 5dx = arctan(x+ 2) + C
37)
∫ 2
0
1
(3 + 5x)2dx
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:∫ 2
0
1
(3 + 5x)2dx =
∫ 13
3
1
5u2du =
−1
5u|133 = − 1
65+
1
15=
2
39
⇒∫ 2
0
1
(3 + 5x)2dx =
2
39
38)
∫sin(lnu) du
Solution: Using direct substitution with t = lnu, that is, u = et, and du = et dt, wehave that: ∫
sin(lnu) du =
∫et sin t dt
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MATH 105 921 Solutions to Integration Exercises
Using integration by parts twice to compute the integral on the right hand side (seethe solution of question 18 for details), we have that:∫
et sin t dt =1
2et sin t− 1
2et cos t+ C
Therefore,∫sin(lnu) du =
1
2elnu sin(lnu)− 1
2elnu cos(lnu) + C =
1
2u sin(lnu)− 1
2u cos(lnu) + C
39)
∫r(ln r)2 dr
Solution: Using integration by parts with u = (ln r)2, du =2 ln r
rdr, and dv = r dr,
v =r2
2, we get that:
∫r(ln r)2 dr =
r2(ln r)2
2−∫r ln r dr
Using integration by parts again on the remaining integral with u1 = ln r, du1 =1
rdr,
and dv1 = r dr, v1 =r2
2, we get that:
∫r ln r dr =
r2 ln r
2−∫
r
2dr =
r2 ln r
2− r2
4+ C
Therefore, ∫r(ln r)2 dr =
r2(ln r)2
2− r2 ln r
2+r2
4+ C
40)
∫1
x3 − xdx
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MATH 105 921 Solutions to Integration Exercises
Solution: Using partial fraction, we get:
1
x3 − x=A
x+
B
x+ 1+
C
x− 1=A(x2 − 1) +B(x2 − x) + C(x2 + x)
x3 − x
=(A+B + C)x2 + (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 = 1
2and B = −1
2. So,∫
1
x3 − xdx =
∫−1
xdx−
∫1
2(x+ 1)dx+
∫1
2(x− 1)dx
⇒∫
1
x3 − xdx = − ln |x| − 1
2ln |x+ 1|+ 1
2ln |x− 1|+ C
Remark: This involves partial fractions with 3 distinct roots in the denominator, whichyou are not required to master in this course!
41)
∫sec3 u du
Solution: We use the reduction formula:∫sec3 u du =
1
2tanu secu+
1
2
∫secu du =
1
2tanu secu+
1
2ln | secu+ tanu|+ C
42)
∫ √x2 − 2x− 8
x− 1dx
Solution: Observe that x2 − 2x − 8 = (x − 1)2 − 9. Using direct substitution witht = x− 1, and dt = dx, we get:∫ √
x2 − 2x− 8
x− 1dx =
∫ √t2 − 9
tdt
Using inverse trigonometric substitution with t = 3 sec y, and dt = 3 sec y tan y dy, we
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MATH 105 921 Solutions to Integration Exercises
get: ∫ √t2 − 9
tdt =
∫ √9 sec2 y − 9
3 sec y3 sec y tan y dy =
∫3 tan2 y dy
=
∫3(sec2 y − 1) dy = 3 tan y − 3y + C
⇒∫ √
x2 − 2x− 8
x− 1dx = 3 tan(arccos
(3
t
))− 3 arccos
(3
t
)+ C =
√t2 − 9− 3 arccos
(3
t
)+ C
=√
(x− 1)2 − 9− 3 arccos
(3
x− 1
)+ C
43)
∫ √r2 − 1
rdr
Solution: Using inverse trigonometric substitution with sec s = r, that is, s =
arccos
(1
r
), and sec s tan s ds = dr, we get:
∫ √r2 − 1
rdr =
∫ √sec2 s− 1
sec ssec tan s ds =
∫tan2 s ds
=
∫(sec2 s− 1) ds = tan s− s+ C
⇒∫ √
r2 − 1
rdr = tan(arccos
(1
r
))− arccos
(1
r
)+ C =
√r2 − 1− arccos
(1
r
)+ C
44)
∫(et
2
+ 16)tet2
dt
Solution: Using direct substitution with u = et2
and du = 2tet2dt, we get:∫
(et2
+ 16)tet2
dt =
∫1
2(u+ 16) du =
1
4u2 + 8u+ C
⇒∫
(et2
+ 16)tet2
dt =1
4e2t
2
+ 8et2
+ C
45)
∫√y ln y dy
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MATH 105 921 Solutions to Integration Exercises
Solution: Using integration by parts with u = ln y, du =1
ydy and dv =
√y dy,
v =2
3y
32 , we get:∫
√y ln y dy =
2
3y
32 ln y −
∫2
3y
12 dy =
2
3y
32 ln y − 4
9y
32 + C
46)
∫cos θ
1 + sin2 θdθ
Solution: Using direct substitution with u = sin θ, and du = cos θ dθ, we have that:∫cos θ
1 + sin2 θdθ =
∫1
1 + u2du = arctanu+ C
⇒∫
cos θ
1 + sin2 θdθ = arctan(sin θ) + C
47)
∫1
x2√x2 + 4
dx
Solution: Using inverse trigonometric substitution with 2 tanu = x, and 2 secu du =dx, we get:∫
1
x2√x2 + 4
dx =
∫1
4 tan2 u√
4 tan2 u+ 42 sec2 u du =
∫2 sec2 u
8 tan2 u secudu
=
∫cos2 u
4 cosu sin2 udu =
∫1
4cotu cscu du
= −1
4cscu+ C = −1
4csc(arctan
(x2
)) + C
⇒∫
1
x2√x2 + 4
dx = −√x2 + 4
4x+ C
48)
∫tet
2
dt
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MATH 105 921 Solutions to Integration Exercises
Solution: Using direct substitution with u = t2, and du = 2t dt, we have that:∫tet
2
dt =
∫1
2eu du =
1
2eu + C
⇒∫tet
2
dt =1
2et
2
+ C
49)
∫cos(πt) cos(sin(πt)) dt
Solution: Using direct substitution with u = sin(πt) dt, and du = π cos(πt) dt, wehave that: ∫
cos(πt) cos(sin(πt)) dt =
∫1
πcosu du =
1
πsinu+ C
⇒∫
cos(πt) cos(sin(πt)) dt =1
πsin(cos(πt)) + C
50)
∫ π4
0
sin5(x) dx
Solution: Using direct substitution with u = cosx, and du = − sinx dx, when x = 0,
then u = 1, and when x =π
4, u =
1√2
. We have that:
∫ π4
0
sin5(x) dx =
∫ π4
0
(sin2(x))2 sinx dx =
∫ π4
0
(1− cos2 x)2 sinx dx
=
∫ 1√2
1
−(1− u2)2 du =
∫ 1√2
1
(−1 + 2u2 − u4) du
= (−u+2
3u3 − 1
5u5) |
1√2
1
= (− 1√2
+1
3√
2− 1
20√
2)− (−1 +
2
3− 1
5) = − 43
60√
2+
8
15
⇒∫ π
4
0
sin5(x) dx = − 43
60√
2+
8
15
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