Arts Sci 1D06 Solutions to Assignment #9
1.
2.
3.
4.
5.
6.
Z
7. (a)
sin4 (x) cos2 (x)dx
Use the identities sin2 (x) = 21 (1 − cos(2x)) and cos2 (x) = 12 (1 +
cos(2x)) to get the integral
Z
1
(1 − cos(2x))2 (1 + cos(2x))dx =
8
Z
1
(1 − 2 cos(2x) + cos2 (2x))(1 + cos(2x))dx =
8
Z
1
1 − cos(2x) − cos2 (2x)) + cos3 (2x)dx =
8
Z
Z
1
1
2
3
x − sin(2x) − cos (2x)dx + cos (2x)dx =
8
2
Z
Z
1
1
1
2
(1 + cos(4x))dx + (1 − sin (2x)) cos(2x)dx =
x − sin(2x) −
8
2
2
Z
1
1
1
1
2
x − sin(2x) −
x + sin(4x) + (1 − sin (2x)) cos(2x)dx .
8
2
2
4
To compute the last integral,
R make the substitution u = sin(2x).
Then du = 2 cos(2x)dx and (1 − sin2 (2x)) cos(2x)dx becomes
Z
1
1
1 3
1
1
2
3
(1 − u )du =
u− u =
sin(2x) − sin(2x) .
2
2
3
2
3
Putting this all together, we get that the integral is
1
1
1
1
1 3
1
x − sin(2x) − (x + sin(4x)) +
sin(2x) − sin (2x)
=
8
2
2
4
2
3
1 1
1
1 3
x − sin(4x) − sin (2x) + C =
8 2
8
6
1
1
1 3
x − sin(4x) − sin (2x) + C
16
4
3
Z
(b)
sin3 (x) cos−2 (x)dx Use the substitution u = cos(x). Then du =
− sin(x)dx and the integral becomes
Z
Z
Z
(1 − u2 )
2
−2
2 −2
− sin(x)(1 − cos (x)) cos (x)dx = − (1 − u )u du = −
du =
u2
Z
1
− (u−2 − 1)du = u−1 + u + C =
+ cos(x) + C = cos(x) + sec(x) + C.
cos(x)
Z
(c)
1
dx Use the trig substitution x = tan(u). Then dx =
(1 + x2 )2
sec2 (u)du and the integral becomes
Z
Z
1
sec2 (u)
2
du = cos2 (u)du =
sec (u)du =
(1 + tan2 (u))2
sec4 (u)
Z
1
1
1
u + sin(2u) + C.
(1 + cos(2u))du =
2
2
2
Z
To express this in terms of x, use that u = arctan(x) and 12 sin(2u) =
x
sin(u) cos(u). If tan(u) = x, then sin(u) = √1+x
2 and cos(u) =
x
√ 1
so sin(u) cos(u) = 1+x2 . Thus the integral is
1+x2
1
2
x
arctan(x) +
1 + x2
+ C.
3/2
Z
(d)
1
dx
(9 − x2 )3/2
0
Use the trig substitution x = 3 sin(u). Then dx = 3 cos(u)du and
the lower and upper limits of integration become arcsin(0/3) = 0
and arcsin(1/2) = π/6 respectively. Then the integral becomes
π/6
Z
0
1
9
Z
0
1
1
3
cos(u)du
=
2
9
(9 − 9 sin (u))3/2
π/6
sec2 (u)du =
Z
0
π/6
cos(u)
du =
(cos2 (u))3/2
1
1
1
π/6
tan(u)|0 = (tan(π/6) − tan(0)) = √ .
9
9
9 3