Math 220 November 8 I. Evaluate thel indefinite integral. 1. ∫ (e3x +

Math 220
November 8
I. Evaluate thel indefinite integral.
1.
Z
(e3x + sin(3x) + tan(3x))dx
2.
Z
3.
4.
Z
Z
Z
6.
x2
dx
x3 + 5
√
tan( x)
√
dx
x
Z
Z
sin(x)
dx
cos(x)
csc2 (x) cot(x))dx
5.
7.
(x cos(x2 )dx
x2 sin(x3 + 1)dx
8.
Z
x
dx
1 + x2
9.
Z
x
dx
1 + 16x4
10.
Z
√
x 3 − xdx
11.
Z
√
x3 x2 − 1dx
II. Evaluate the integral.
1
1.
1
Z
2
xe−x dx
0
2.
π
Z
x3 sin(x)dx
−π
3.
1
x
dx
1 + x4
1
√
t t + 1dt
Z
0
4.
Z
0
5.
Z
π/4
tan(x)dx
0
6.
Z
π/2
sin(2x) + cos(4x)dx
0
2
1
Solutions
I. Evaluate thel indefinite integral.
1.
Z
(e3x + sin(3x) + tan(3x))dx
Answer:
Z
(e3x + sin(3x) + tan(3x))dx =
u = 3x
du = 3dx
1
du = dx
3
Z
1
(eu + sin(u) + tan(u)) du
3
Z
Z
1
1
=
(eu + sin(u))du +
tan(u)du
3
3
Z
1 u
1
sin(u)
= (e − cos(u)) +
du
3
3
cos(u)
=
w = cos(u)
dw = − sin(u)du
=
=
=
=
2.
Z
1
1 u
−1
(e − cos(u)) +
dw
3
3
w
1 u
1
(e − cos(u)) + (− ln |w|) + C
3
3
1
1 u
(e − cos(u)) + (− ln | cos(u)|) + C
3
3
1 3x
−1
(e − cos(3x)) +
(ln | cos(3x)|) + C
3
3
Z
(x cos(x2 )dx
3
Answer:
Z
(x cos(x2 )dx =
u = x2
du = 2xdx
Z
=
1
cos(u) du
2
1
sin(u) + C
2
1
= sin(x2 ) + C
2
=
3.
Z
sin(x)
dx
cos(x)
Answer:
Z
sin(x)
dx =
cos(x)
u = cos(x)
du = − sin(x)dx
−1
du
u
= − ln |u| + C
= − ln | cos(x)| + C
Z
=
4.
Z
csc2 (x) cot(x))dx
4
Answer:
Z
Z
2
csc (x) cot(x))dx =
csc(x)(csc(x) cot(x))dx
u = csc(x)
du = − csc(x) cot(x)dx
Z
=
u(−du)
Z
= − udu
−u2
+C
2
− csc2 (x)
+C
=
2
=
5.
Z
x2
dx
x3 + 5
Answer:
Z
x2
dx =
x3 + 5
u = x3 + 5
du = 3x2 dx
Z
=
11
du
u3
1
ln |u| + C
3
ln |x3 + 5|
=
+C
3
=
6.
Z
√
tan( x)
√
dx
x
5
Answer:
Z
√
tan( x)
√
dx =
x
u=
√
x
1
du = √ dx
2 x
Z
=
tan(u)2du
= −2 ln | cos(u)| + C
√
= −2 ln | cos( x)| + C
7.
Z
x2 sin(x3 + 1)dx
Answer:
Z
x2 sin(x3 + 1)dx =
u = x3 + 1
du = 3x2 dx
Z
1
sin(u) du
3
−1
cos(u) + C
=
3
−1
=
cos(x3 + 1) + C
3
=
8.
Z
x
dx
1 + x2
Answer:
Z
x
dx =
1 + x2
u = 1 + x2
du = 2xdx
Z
=
11
du
u2
1
ln |u| + C
2
ln |1 + x2 |
=
+C
2
=
6
9.
Z
x
dx
1 + 16x4
Answer:
Z
x
dx =
1 + 16x4
=
=
Z
u = 4x2
du = 8xdx
Z
=
x
dx
1 + (4x2 )2
1 1
du
1 + u2 8
1
tan−1 (u) + C
8
1
= tan−1 (4x2 ) + C
8
=
10.
Z
√
x 3 − xdx
Answer:
Z
√
x 3 − xdx =
u=3−x
x=3−u
du = −dx
Z
=
Z
=
√
(3 − u) udu
(3u1/2 − u3/2 )du
u3/2 u5/2
−
+C
3/2
5/2
2
= 2u3/2 − u5/2 + C
5
2
= 2(3 − x)3/2 − (3 − x)5/2 + C
5
=3
11.
Z
√
x3 x2 − 1dx
7
Answer:
Z
x
3
√
Z
x2 − 1dx =
√
x2 x2 − 1xdx
u = x2 − 1
x2 = u − 1
du = 2xdx
√ 1
(u − 1) u du
2
Z
1
=
u3/2 − u1/2 du
2
1 u5/2 u3/2
−
)+C
= (
2 5/2
3/2
1
1
= (x2 − 1)5/2 − (x2 − 1)3/2 + C
5
3
Z
=
II. Evaluate the integral.
1.
Z
1
2
xe−x dx
0
Answer:
Z
1
2
xe−x dx =
u = −x2
0
du = −2xdx
Z
u(1)
=
eu
u(0)
=
=
=
=
2.
−1
du
2
Z
−1 −1 u
e du
2 0
−1 −1
(e − e0 )
2
−1 −1
(e − 1)
2
1
1
−
2 2e
Z
π
x3 sin(x)dx
−π
8
Answer:
Z
π
x3 sin(x)dx = 0
−π
Since x3 sin(x) is odd.
3.
1
Z
0
x
dx
1 + x4
Answer:
Z
0
1
x
dx =
1 + x4
Z
0
1
x
dx
1 + (x2 )2
u = x2
du = 2xdx
Z
u(1)
=
u(0)
Z 1
1 1
dx
1 + u2 2
1
1
dx
2 0 1 + u2
1
= (tan−1 (1) − tan−1 (0))
2
1 π
= ( − 0)
2 4
π
=
8
=
4.
Z
1
√
t t + 1dt
0
9
Answer:
Z
1
√
t t + 1dt =
u=t+1
0
t=u−1
du = dt
Z
u(1)
=
√
(u − 1) udu
u(0)
2
Z
=
u3/2 − u1/2 du
1
u5/2 u3/2 i2
−
=
5/2
3/2 1
2√
2√
2 2
=(
32 −
8) − ( − )
5
3
5 3
2 √
2 √
= ( 32 − 1) − ( 8 − 1)
5
3
5.
Z
π/4
tan(x)dx
0
Answer:
Z
π/2
iπ/4
tan(x)dx = − ln(cos(x))
0
0
= − ln(cos(π/4)) − (− ln(cos(0))
√
= − ln(1/ 2) + ln(1)
= − ln(2−1/2 )
ln(2)
=
2
6.
Z
π/2
sin(2x) + cos(4x)dx
0
10
Answer:
Z
π/2
Z
π/2
sin(2x) + cos(4x)dx =
0
Z
sin(2x)dx +
0
π/2
cos(4x)dx
0
iπ/2 1
iπ/2
1
= (− cos(2x))
+ sin(4x)
2
4
0
0
1
1
= (− cos(π) + cos(0)) + (sin(2π) − sin(0))
2
4
1
1
= (1 + 1) + (0 − 0)
2
4
=1
11