MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
MATH 2260
Assignment 1
Winter 2014
SOLUTIONS
[5]
1. (a) We use integration by parts, with w = ln(x) so dw = x1 dx and dv = x2 so v = 13 x3 . Thus
we can write
Z
Z
1 3
1
1
2
x ln(x) dx = x ln(x) −
x3 · dx
3
3
x
Z
1
1
x2 dx
= x3 ln(x) −
3
3
1 3
1 1 3
= x ln(x) −
x +C
3
3 3
1
1
= x3 ln(x) − x3 + C.
3
9
[5]
(b) We use integration by parts again, this time letting w = x2 so dw = 2x dx and dv =
sin(x) dx so v = − cos(x). Thus we have
Z
Z
2
2
x sin(x) dx = −x cos(x) + 2 x cos(x) dx.
Now we need to use integration by parts again, with w = x so dw = dx and dv = cos(x) dx
so v = sin(x). Hence
Z
Z
2
2
x sin(x) dx = −x cos(x) + 2 x sin(x) − sin(x) dx
= −x2 cos(x) + 2x sin(x) − 2[− cos(x)] + C
= −x2 cos(x) + 2x sin(x) + 2 cos(x) + C.
[5]
(c) We can write
5
4x + 4x + 10 = 4 x + x +
2
1
9
2
=4 x +x+
+
4
4
2
1
=4 x+
+9
2
2
2
= (2x + 1)2 + 9.
–2–
Thus we can let u = 2x + 1 so 21 du = dx, and the integral becomes
Z
Z
1
1
dx =
dx
2
4x + 4x + 10
(2x + 1)2 + 9
Z
1
1
du
=
2
2
u +9
u
1 1
= · arctan
+C
2 3
3
1
2x + 1
= arctan
+ C.
6
3
[5]
(d) We can rewrite the given integral as
Z
Z
5
2
5 ln(tan(y))
sec (y)e
dy = sec2 (y)eln(tan (y)) dy
Z
= tan5 (y) sec2 (y) dy.
Now we let u = tan(y) so du = sec2 (y) dy. The integral becomes
Z
Z
2
5 ln(tan(y))
sec (y)e
dy = u5 du
1
= u6 + C
6
1
= tan6 (y) + C.
6
[5]
(e) First note that
x3 − 4x2 − 16x + 64 = x2 (x − 4) − 16(x − 4) = (x − 4)(x2 − 16) = (x + 4)(x − 4)2 .
Now we can decompose the integrand into partial fractions:
A
B
D
−1
1
3
11x − 20
=
+
+
=
+
+
.
2
2
(x + 4)(x − 4)
x + 4 x − 4 (x − 4)
x + 4 x − 4 (x − 4)2
Hence the integral becomes
−1
1
3
+
+
dx
x + 4 x − 4 (x − 4)2
3
= − ln|x + 4| + ln|x − 4| −
+ C.
x−4
11x − 20
dx =
3
x − 4x2 − 16x + 64
[5]
(f) Let u = t3 so
1
3
Z du = t2 dt. The integral becomes
Z
Z
Z
1
5 t3
3 t3
2
t e dt = t e · t dt =
ueu du.
3
–3–
Now we use integration by parts with w = u so dw = du and dv = eu du so v = eu . Then
we have
Z
Z
1
u
u
5 t3
ue − e du
t e dt =
3
1
1
= ueu − eu + C
3
3
1
1 3
3
= t3 et − et + C.
3
3
[5]
(g) Again we decompose the integrand into partial fractions:
2x3 + 8x + 4
A B
Dx + E
2
1
1
2x3 + 8x + 4
=
= + 2+ 2
= + 2− 2
.
4
2
2
2
x + 4x
x (x + 4)
x x
x +4
x x
x +4
Thus
Z
[5]
2x3 + 8x + 4
dx =
x4 + 4x2
Z 1
1
2
+
−
dx
x x2 x2 + 4
x
1 1
+ C.
= 2 ln|x| − − arctan
x 2
2
(h) We use integration by parts with w = e2x so dw = 2e2x dx and dv = cosh(3x) dx so
v = 13 sinh(3x). Then
Z
Z
1 2x
2
2x
e cosh(3x) dx = e sinh(3x) −
e2x sinh(3x) dx.
3
3
Now we use integration by parts a second time, again with w = e2x so dw = 2e2x dx but
this time with dv = sinh(3x) dx so v = 13 cosh(3x). We have
Z
Z
1 2x
2 1 2x
2
2x
2x
e cosh(3x) dx = e sinh(3x) −
e cosh(3x) −
e cosh(3x) dx
3
3 3
3
Z
1 2x
2 2x
4
= e sinh(3x) − e cosh(3x) +
e2x cosh(3x) dx
3
9
9
Z
5
1
2
e2x cosh(3x) dx = e2x sinh(3x) − e2x cosh(3x) + C
9
3
9
Z
3
2
e2x cosh(3x) dx = e2x sinh(3x) − e2x cosh(3x) + C.
5
5