MATH 101 V2A
January 23rd – Practice Problems
Hints and Solutions:
1. Evaluate the following integrals.
Z 1
(a) (i)
sin5 (t) cos6 (t)dt.
5
Solution: See the solutions to the practice problems from January 21st.
Z
1
cos5 (3t)dt.
(ii)
5
Hint: See the solutions to the practice problems from January 21st.
Z
1
cos2 (x)dx.
(iii)
0
Hint: See the solutions to the practice problems from January 21st.
Z
(b)
(i)
0
1
1
dx.
1 + x2
√
Solution: Identify 1 + x2 as the hypotenuse of a right-angled triangle with sides of length
1 and x. Then x = tan(θ), where θ is the angle of the triangle for which the side of length 1
is the adjacent side, and the side of length x is the opposite side. Then du = sec2 (θ)dθ, and
1
1
π
cos(θ) = √1+x
, so dθ = 1+x
2 dx. Now, if x = 0, then tan(θ) = 0, so θ = 0 (since θ ∈ [0, 2 ]),
2
and if x = 1, then tan(θ) = 1, so θ = π4 . Therefore
1
Z
0
Z
(ii)
0
√
2 3
√
1
dx =
1 + x2
π/4
Z
dθ =
0
π
.
4
x3
dx.
16 − x2
√
Hint: Identify 16 − x2 as one side of a right-angled triangle with hypotenuse 4 and whose
other side is of length x. Then sin(θ) = x4 , where θ is the angle of the triangle for which
√
16 − x2 is the adjacent side and x is the opposite side. Use this substitution to get that
Z
0
√
2 3
√
x3
dx =
16 − x2
2
Z
π/3
64 sin3 (θ)dθ.
0
Now use the trigonometric identity sin (θ) + cos2 (θ) = 1 to write sin3 (θ) = sin(θ)(1 − cos2 (θ))
and to break the integral into two integrals. Answer: 40
3 .