(1) Compute the following integrals (a) ∫ x2(x3 + 1)6 dx (b) ∫ ex

(1) Compute the following integrals (a) ∫ x2(x3 + 1)6 dx (b) ∫ ex

(1) Compute the following integrals
Z
Z
√
2 3
6
(a)
x (x + 1) dx
(b)
ex 1 + 2ex dx
Z
1
(d)
0
x2
dx
1 − x3
Z
(g)
Z
(e)
√
Z
(c)
1
dx
x ln(x)
Z
3x + 1dx
(f)
cos(x) sin5 (x) dx
Z
sec2 (x)
p
dx
2 + tan(x)
R
R
Solution: (a) Let u = x3 + 1, du = 3x2 dx. Then I = x2 (x3 + 1)6 dx = u6 (du/3) =
3
dx
4 + 5x
(h)
(1/3)[u7 /7] = (1/21)(x3 + 1)7 + C
R x√
R
(b) Let u = 1 + 2ex , du = 2ex dx. Then I =
e 1 + 2ex dx = u1/2 (du/2) =
R
(1/2) u1/2 du = (1/2)U 3/2 /(3/2) = (1/3)(1 + 2E X )3/2 + C
R 1
R
(c) u = ln x, du = (1/x)dx. Then I = x ln(x)
dx = (1/u)du = ln u = ln(ln x) + C
R
(d) u = 1 − x3 , du = −3x2 dx. Then I = (1/u)(du/ − 3) = (−1/3) (1/u)du =
−(1/3) ln u = −(1/3) ln(1 − x3 ) + C
R
u1/2 (du/3) = u3/2 /(9/2) = (2/9)(3x + 1)3/2 + C
R
(f) u = sin x, du = cos x dx. Then I = u5 du = u6 /6 = sin6 x/6 + C
R
(g) u = 4 + 5x, du = 5dx. Then I = 3 (1/u)(du/5) = (3/5) ln u = (3/5) ln(4 + 5x) + C
R
R
(h) u = 2 + tan(x), du = sec2 x dx. So I = (1/u1/2 )du = u−1/2 du = u1/2 /(1/2)
√
= 2 u = 2(2 + tan x)1/2
R
(2) The integral sin(x) cos(x)dx can be done in several ways:
(e) u = 3x + 1, du = 3dx. Then I =
(a) using the substitution u = sin(x) (or u = cos(x)) OR
(b) by first using the trigonometric identity sin(2x) = 2 sin(x) cos(x) and then integrating.
Show that the two answers are equivalent.
Solution: (a) u = sin(x), du = cos(x)dx, so I =
R
sin x cos x dx =
R
udu = u2 /2 =
sin2 x/2 + C.
(b) sin(2x) = 2 sin x cos x (trig identity). So I =
R
(1/2) sin(u)(du/2) = (−1/4) cos(2x) + C.
R
sin(x) cos(x)dx = (1/2) sin(2x)dx =
The identity cos(2x) = cos2 x − sin2 x can be used to show that the two results are equivalent: I = (−1/4)(1 − 2 sin2 x) + C = sin2 x/2 + (C − 1/4). The quantity at the end is just
another constant.
1
(3) The integral
R1 √
0
1 − x2 dx can be done by making the trigonometric substitution
x = sin(u) and dx = cos(u)du.
(a) Show that this reduces the integral to
R
cos2 (u) du.
(b) Now use the identity cos2 (u) = [1 + cos(2u)]/2 to express the integral in a simpler
form. Then integrate.
(c) Explain why the answer is the same as the area of 1/4 of a circle of radius 1.
p
Solution: (a) Let x = sin(u), dx = cos(u)du. Then I = 1 − sin2 u(cos u du). But
R
R
1 − sin2 u = cos2 u, so that I = cos u cos u du = cos2 u du.
R
(b) Using the identity we get I = (1/2) [1 + cos(2u)]du = (1/2)[u + (1/2) sin(2u)] =
(1/2)[u + sin(u) cos(u)]. Now we need to express in terms of x, so we use x = sin u, so also
p
√
√
1 − x2 = 1 − sin2 u = cos u to rewrite the above as I = (1/2)[sin−1 x + x 1 − x2 ]|10 =
(1/2)(sin−1 1 − sin−1 0 = (1/2)(π/2 − 0) = π/4.
√
(c) The function y = f (x) = 1 − x2 describes part of a circle, and integrating from 0 to
1 gives the area under that curve in the first quadrant, which is 1/4 the area of a circle.
(4) Compute
Z
dx
+ x2
I can be reduced to an inverse tangent type integral by a bit of algebraic rearrangement
I=
a2
and a substitution of the form u = x/a.
Z
Solution:
I=
dx
=
2
a + x2
Z
a2 [1
dx
.
+ (x2 /a2 )]
Substitute u = x/a, then du = dx/a so
Z
Z
adu
du
I=
= (1/a)
= (1/a) tan−1 (u) + C = (1/a) tan−1 (x/a) + C.
2
2
2
a [1 + u ]
[1 + u ]
(5) Integrate the following
Z
Z
Z
1
1
√
(a)
dx
(b)
x √
dx
(c)
cos(3x) (1 − sin(3x)) dx
2 − 3x2
2 + 3x2
Z
Z
Z
2
2
2
5
(f)
t et dt
(d)
sec(x) tan(x) dx
(e)
(x sin (x) + x cos (x)) dx
Solution: (a) Using u = (3/2)x2 , we get
√
Z
6x
1
1
−1
√
)+C
dx = √ sin (
2
2 − 3x2
3
2
(b) Using u = 2 + 3x2 , we get
Z
1
1p
x √
dx =
2 + 3x2 + C
3
2 + 3x2
(c) With the appropraite trig identity
Z
Z
sin(6x)
−1
cos(3x) (1 − sin(3x)) dx =
(cos(3x) −
) dx =
(1 − sin(3x))2 + C
2
6
(d) Again, with a trig identity
Z
Z
sec(x) tan(x) dx =
sin(x)
dx =
cos2 (x)
Z
(−1/u2 )du = (1/u) = 1/ cos(x)
(e) Another trig identity...
Z
Z
2
2
5
(x sin (x) + x cos (x)) dx = (x)5 dx = (1/6)x6 + C
(f) Basic subsitution...
Z
2
2
t et dt = (1/2)et + C
(6) (a) Find the following indefinite integrals using the substitution rule:
Z
(i)
sin(3x) dx
√
Z
cos x
√
(ii)
dx
x
Z
p
(iii)
x3 x4 + 1 dx
Z
4
(iv)
dx .
1 + 2x
(b) Find the following definite integrals using the substitution rule:
Z
5
(i)
√
3 + 2x dx
0
Z
π/4
(ii)
sin(4t) dt
0
Z
(iii)
0
Z
4
dx
(x − 2)3
√
π
x cos(x2 ) dx .
(iv)
0
3
Solution: (a)(i) Set u = 3x. Then du = 3 dx, and
Z
Z
− cos 3x
1
− cos u
=
+ C.
sin(3x) dx =
sin u du =
3
3
3
(ii) Set u =
√
x. Then du = (1/2)x−1/2 dx, and
√
Z
√
cos x
√
dx = 2 cos u du = 2 sin u = 2 sin x + C.
x
Z
(iii) Set u = x4 + 1. Then du = 4x3 dx and
Z
x
3
p
x4
1
+ 1 dx =
4
Z
(x4 + 1)3/2
1 u3/2
=
+ C.
4 3/2
6
u1/2 du =
(iv) Set u = 1 + 2x. Then du = 2 dx, and
Z
4
dx =
1 + 2x
Z
2
du = 2 log |u| = 2 log |1 + 2x| + C.
u
(b)(i) Set u = 3 + 2x. Then du = 2 dx and
Z
0
5
√
1
3 + 2x dx =
2
Z
13
u
1/2
3
1
du =
2
2u3/2
3
13
3
u3/2
=
3
13
= 13.89 .
3
(ii) Set u = 4t. Then du = 4 dt, and
Z
π/4
0
1
sin(4t) dt =
4
Z
π
sin u du =
0
1
1
π
[− cos u]0 = .
4
2
(iii) Set u = x − 2, du = dx. Then formally
Z
4
0
dx
=
(x − 2)3
Z
2
u
−2
−3
u−2
du =
−2
2
.
−2
But to tell the truth this makes no sense, because the denominator x − 2 vanishes in the
interval of integration.
(iv) Set u = x2 , du = 2x dx. Then
Z
√
π
1
x cos(x ) dx =
2
Z
π
2
0
cos u du =
0
1
π
[sin u]0 = 0 .
2
(7) A hot, wet summer is causing a mosquito population explosion in a lake resort area.
2
The number of mosquitos is increasing at an estimated rate of 2000 + 2te0.5t per week
4
(where t is time in weeks). By how much does the mosquito population increase between
the beginning of the fifth and the end of the ninth week of summer?
Solution: The beginning of the fifth week is the end of the fourth. So we get the integral
Z
9
0.5t2
2000 + 2te
4
Z
Z
9
dt =
9
4
2
2te0.5t dt .
2000 dt +
4
Set u = t2 , du = 2t dt. This is
Z
81
10, 000 +
16
81
e0.5u du = 10, 000 + 2e0.5u 16 .
(8) Evaluate
the integral. You must show all your work to get credit.
Z
1
dy
1−y
Solution: Substitute u = 1 − y, du = −dy.
Z
1
dy = − log |1 − y|
1−y
(9) Evaluate the following integral. You must show all your work to get credit.
Z S
k1
dn (k2 outside the range [1, S])
1 k2 − n
Solution: Substitute u = k2 − n, du = −dn:
Z
S
1
k −S
−k1 2
k1
dn =
k2 − n
u k2
(10) Integrate
Z
1
ln(x) dx
(a)
Z x
(b)
tan(x) dx
Z
Z
ln x
u2
(ln x)2
Solution:
dx = u du =
+C =
+ C, where u = ln x.
x
2 Z
2
Z
Z
sin x
−du
(b)
tan x dx =
dx =
= − ln |u| + C = − ln | cos x| + C, where
cos x
u
u = cos x.
5