MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
Basic Integration
Math 1001 Worksheet
Winter 2016
SOLUTIONS
1. (a) We use u-substitution. Let u = x2 − 9 so du = 2x dx and
becomes
Z
Z
1
1
x
√
√ du
dx =
2
u
x2 − 9
Z
1
1
=
u− 2 du
2
1
2
du = x dx. The integral
1
1 u2
= 1 +C
2 2
√
= x2 − 9 + C.
(b) This is an arcsecant type integral with k = 3, so
Z
Z
x
1
1
1
√
√
dx =
dx = arcsec
+ C.
3
3
x x2 − 9
x x2 − 32
(c) We use integration by parts. Let w = x so dw = dx, and let dv = csc2 (9x) dx so
v = − 91 cot(9x). Then
Z
Z
1
1
2
cot(9x) dx
x csc (9x) dx = − x cot(9x) +
9
9
1
1 1
= − x cot(9x) + · ln|sin(9x)| + C
9
9 9
1
1
= − x cot(9x) +
ln|sin(9x)| + C.
9
81
(d) We use u-substitution. Let u = x5 so du = 5x4 dx and
becomes
Z
Z
1
4 x5
x e dx =
eu du
5
1
= eu + C
5
1 5
= ex + C.
5
1
5
du = x4 dx. The integral
–2–
(e) We begin with u-substitution. Let u = x5 so du = 5x4 dx and 15 du = x4 dx. The integral
becomes
Z
Z
Z
1
9 x5
5 x5
4
x e dx = x e (x dx) =
ueu du.
5
Now we use integration by parts, letting w = u so dw = du, and dv = eu du so v = eu .
We obtain
Z
Z
1
u
u
9 x5
ue − e du
x e dx =
5
1
1
= ueu − eu + C
5
5
1
1 5
5
= x5 ex − ex + C.
5
5
(f) We begin by completing the square:
8
4
2
9x − 12x + 8 = 9 x − x +
3
9
4
4
8 4
2
=9 x − x+
+ −
3
9
9 9
"
#
2
2
4
=9 x−
+
3
9
2
= (3x − 2)2 + 4.
Thus the integral becomes
Z
9x2
1
1
dx =
dx.
− 12x + 8
(3x − 2)2 + 4
Now let u = 3x − 2 so du = 3 dx and 31 du = dx. Finally,
Z
Z
1
1
1
dx =
du
2
2
9x − 12x + 8
3
u +4
u
1 1
= · arctan
+C
3 2
2
1
3x − 2
= arctan
+ C.
6
2
(g) We try integration by parts, with w = e4x so dw = 4e4x dx and dv = cos(x) dx so
v = sin(x). Then
Z
Z
4x
4x
e cos(x) dx = e sin(x) − 4 e4x sin(x) dx.
–3–
We try integration by parts a second time. Again, we let w = e4x so dw = 4e4x dx, and
now we let dv = sin(x) dx so v = − cos(x). Thus
Z
Z
4x
4x
4x
4x
e cos(x) dx = e sin(x) − 4 −e cos(x) + 4 e cos(x) dx
4x
Z
4x
= e sin(x) + 4e cos(x) − 16
Z
17
Z
e4x cos(x) dx
e4x cos(x) dx = e4x sin(x) + 4e4x cos(x) + C
e4x cos(x) dx =
4
1 4x
e sin(x) + e4x cos(x) + C.
17
17
(h) We begin by performing long division:
6x − 1
2x − 5 12x2 − 32x + 14
12x2 − 30x
−2x + 14
−2x + 5
9
Now we can write
Z
12x2 − 32x + 14
dx =
2x − 5
Z 6x − 1 +
= 3x2 − x +
(i) Let u = ln(x) so du =
1
x
Z
9
2x − 5
dx
9
ln|2x − 5| + C.
2
dx. The integral becomes
Z
1
1
q
√
dx =
du
2
2
4
−
u
x 4 − ln (x)
u
= arcsin
+C
2
ln(x)
= arcsin
+ C.
2
(j) Recall that 1 + tan2 (x) = sec2 (x), so
Z
Z
2
2
2
2
cos (x)[1 + tan (x)] dx = cos (x) sec (x) dx = dx = x + C.
Alternatively, we could write
Z
Z
Z
2
2
2
2
cos (x)[1 + tan (x)] dx = [cos (x) + sin (x)] dx = dx = x + C.