MATH 105 921 SOLUTIONS TO INTEGRATION EXERCISES COMPLETE TEST QUESTIONS AND VERIFIED ANSWERS 2024

MATH 105 921 Solutions to Integration Exercises

MATH 105 921 Solutions to Integration Exercises
s2 + 1
Z
1) ds
s2 − 1

Solution: Performing polynomial long division, we have that:
Z 2
s +1 2
Z
ds = (1 + ) ds
s2 − 1 s2 − 1
2
Z Z
= ds + 2
ds
s −1
2
Z
=s+ 2
ds
s −1
Using partial fraction on the remaining integral, we get:

2 A B A(s + 1) + B(s − 1) (A + B)s + (A − B)
= + = =
s2 −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:
2 1 1
Z Z Z
2
ds = ds − ds
s −1 s−1 s+1
Therefore,

s2 + 1 2
Z Z
ds = s + ds
s2 − 1 s2 − 1
1 1
Z Z
=s+ ds − ds
s−1 s+1
= s + ln |s − 1| − ln |s + 1| + C


Z 0 √
2) x 1 + 2x dx
4



Solution: Using direct substitution with u = 1 + 2x and du = 2dx, we may write




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,MATH 105 921 Solutions to Integration Exercises


x = 21 (u − 1). Moreover, when x = 4, u = 9, and when x = 0, u = 1. Thus,
0 √ 1 √
1
Z Z
x 1 + 2x dx = (u − 1) u du
4 4
Z9 1
1 3 1
= (u 2 − u 2 ) du
9 4
1 5 1 3
= ( u 2 − u 2 ) |19
10 6
1 1 243 27
=( − )−( − )
10 6 10 6
−298
=
15


Z
3) sin2 x cos2 x dx


1−cos(2x) 1+cos(2x)
Solution: Using half-angle identities sin2 x = 2
and cos2 x = 2
, we get:

1
Z Z
2 2
sin x cos x dx = (1 − cos(2x))(1 + cos(2x)) dx
4
1
Z
= (1 − cos2 (2x)) dx
4
1 1
Z Z
= dx − cos2 (2x) dx
4 4
x 1
Z
= − cos2 (2x) dx
4 4
1+cos(4x)
On the remaining integral, we apply the half-angle identity cos2 (2x) = 2
, and
obtain:
1 + cos(4x) x 1
Z Z
2
cos (2x) dx = dx = + sin(4x) + C
2 2 8
Hence,
x 1 x 1 1
Z
x
sin2 x cos2 x dx = − ( + sin(4x)) + C = − sin(4x) + C
4 4 2 8 8 32


√
Z
4) sin( w) dw



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, MATH 105 921 Solutions to Integration Exercises


√ 1
Solution: Using direct substitution with t = w, and dt = √
2 w
dw, that is, dw =
√
2 w dt = 2t dt, we get:
√
Z Z
sin( w) dw = 2t sin t dt

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
sin( w) dw = −2 w cos( w) + 2 sin( w) + C



ln(x)
Z
5) dx
x

1
Solution: Using direct substitution with u = ln(x) and du = x
dx, we get:

ln(x) u2
Z Z
dx = u du = +C
x 2
ln(x) 1
Z
⇒ dx = (ln(x))2 + C
x 2

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 sin t cos t dt − sin t dt = 2 sin t cos2 t dt + cos t
2



On the remaining integral, using direct substitution with u = cos t and du = − sin t dt,
we have that:
2 2
Z Z
2 sin t cos t dt = −2u2 du = − u3 + C = − cos3 t + C
2
3 3



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