MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
Assignment 2
Mathematics 1001
Winter 2016
SOLUTIONS
[5]
1. (a) Let u =
1
1
1
so du = − 2 dx and − du = 2 dx. The integral becomes
x
x
x
Z 1
Z
ex
dx = − eu du
x2
= −eu + C
1
= −e x + C.
[5]
(b) Let u =
√
√
2
3
1
3t so du = √ dt and du = √ dt. The integral becomes
3
2 t
3t
√
Z
Z
6 sec2 3t
2
√
dt = · 6 sec2 (u) du
3
3t
= 4 tan(u) + C
√ = 4 tan
3t + C.
[5]
(c) Let u =
π
− cos(x) so du = sin(x) dx. The integral becomes
7
Z
Z
π
sin(x) sin
− cos(x) dx = sin(u) du
7
= − cos(u) + C
π
− cos(x) + C.
= − cos
7
[5]
1
(d) Let u = 3 − x2 so du = −2x dx and − du = x dx. Furthermore, x2 = 3 − u so
2
x4 = (3 − u)2 . The integral becomes
Z
Z
1
5
2 7
x (3 − x ) dx = −
x4 u7 du
2
Z
1
=−
(3 − u)2 u7 du
2
Z
1
=−
(9u7 − 6u8 + u9 ) du
2
1 9 8 2 9
1 10
=−
u − u + u
+C
2 8
3
10
1
1
9
= − (3 − x2 )8 + (3 − x2 )9 − (3 − x2 )10 + C.
16
3
20
–2–
[5]
(e) Observe that we can rewrite the given integral as
Z
Z
Z
1
3x−1
x
3x −1
x
8x tan(8x ) dx.
2
tan(8 ) dx = 2 2 tan(8 ) dx =
2
Now we let u = 8x so du = 8x ln(8) dx and
Z
23x−1 tan(8x ) dx =
1
ln(8)
du = 8x dx. The integral becomes
Z
1
tan(u) du
2 ln(8)
1
[− ln|cos(u)|] + C
=
2 ln(8)
=−
[5]
[5]
ln|cos(8x )|
+ C.
ln(64)
2. (a) We use u-substitution with u = 5x3 − 6x2 so du = (15x2 − 12x) dx and
4x) dx. Now we can write
Z
Z
5x2 − 4x
10x2 − 8x
dx
=
2
dx
5x3 − 6x2
5x3 − 6x2
Z
1
2
du
=
3
u
2
= ln|u| + C
3
2
= ln|5x3 − 6x2 | + C.
3
(b) We begin by performing long division:
2x2 − 7x
3x + 1 6x − 19x2 − 7x + 1
6x3 + 2x2
− 21x2 − 7x + 1
− 21x2 − 7x
1
3
Thus we have
Z
6x3 − 19x2 − 7x + 1
dx =
3x + 1
Z 2x2 − 7x +
1
3x + 1
dx
2
7
1
= x3 − x2 + ln|3x + 1| + C.
3
2
3
[5]
(c) We begin by performing long division:
3x
x3 + 2 3x4 − x2 + 6x
3x4
+ 6x
2
−x
1
3
du = (5x2 −
–3–
Thus we can write
Z Z
Z
x2
3x4 − x2 + 6x
x2
3 2
dx
=
3x
−
dx
=
dx.
x
−
x3 + 2
x3 + 2
2
x3 + 2
For the remaining integral, let u = x3 + 2 so du = 3x2 dx and
Z
3x4 − x2 + 6x
3
dx = x2 −
3
x +2
2
3
= x2 −
2
3
= x2 −
2
1
3
du = x2 dx. Then
Z
1
1
du
3
u
1
ln|u| + C
3
1
ln|x3 + 2| + C.
3