MAT 21B: Winter 2017
8.1.2
x2
x2 +1
R
Z
√
dx
√ =
x− x
x − 1 and du =
R3
4x2 −7
−1 2x+3
8.1.12
Z 2
x2 + 1 − 1
x +1
1
dx =
dx
+
x2 + 1
x2+1 x2 + 1
Z
1
= 1+
dx = x + arctan x + C
1 + x2
x2
dx =
x2 + 1
Z
dx
√
x− x
R
1 :u=
Due February 17
dx
Z
8.1.6
Homework
Z
1
dx
1
√ √
=
x x−1
Z
√
2du
= 2 ln |u| + C = 2 ln | x − 1| + C
u
dx
√
2 x
dx
3
Z
−1
Z 3 2
4x2 − 9 + 2
4x − 9
2
dx =
+
dx
2x + 3
2x + 3
2x + 3
−1
−1
Z 3
(2x − 3)
(2x
+
3)
2
=
+
dx
2x+
3
2x + 3
−1
3
2
= x − 3x + ln |2x + 3| = ln(9) − 4
4x2 − 7
dx =
2x + 3
Z
3
−1
8.1.23
R 0.5π √
0
1 − cos θ dθ
Z
Z
√
Z r
Z r
1 + cos θ
1 − cos2 θ
1 − cos θ dθ =
(1 − cos θ)
dθ =
dθ
1 + cos θ
1 + cos θ
s
Z
Z
sin θ
sin2 θ
√
dθ =
dθ
=
1 + cos θ
1 + cos θ
Z
√
√
1
−du
√ = −2 u + C = −2 1 + cos θ + C
=
u
0.5π
√
0.5π
√
√
1 − cos θ dθ = −2 1 + cos θ
= −2 2 + 2
∴
0
0
1 : u = 1 + cos θ and du = − sin θ dθ
8.1.39
R
dx
1+ex
Z
dx
=
1 + ex
Z
dx
e−x
·
=
1 + ex e−x
e−x
dx = − ln |e−x + 1| + C
e−x + 1
Z
1
MAT 21B: Winter 2017
8.2.1
R
x sin x2 dx
Z
Homework
1
x
x
x sin dx = −2x cos −
2
2
Z
−2 cos
Due February 17
x
x
x
dx = −2x cos + 4 sin + C
2
2
2
1 : u = x, du = 1, v = −2 cos x2 , and dv = sin x2
8.2.4
R
x2 sin x dx
Z
1
x sin x dx = −x2 cos x −
2
Z
−2x cos x dx
Z
2
2
= −x cos x − −2x sin x − −2 sin x dx
= −x2 cos x + 2x sin x + 2 cos x + C
1 : u1 = x2 , du1 = 2x, v1 = − cos x, and dv1 = sin x
2 : u2 = −2x, du2 = −2, v2 = sin x, and dv2 = cos x
8.2.8
R
xe3x dx
Z
xe
1 : u = x, du = 1, v =
e3x
,
3
3x
1 e3x
−
dx = x
3
Z
e3x e3x
e3x
dx = x
−
+C
3
3
9
and dv = e3x
R
8.2.16 p4 e−p dp
Following the example 7 on page 466,
Z
p4 e−p dp = e−p −p4 − 4p3 − 12p2 − 24p − 24 + C
8.2.24
R
e−2x sin (2x) dx
Z
Z
1 1 −2x
−2x
e
sin (2x) dx = − e
cos (2x) − e−2x cos (2x)dx
2
Z
1
2 1 −2x
−2x
=− e
cos (2x) −
e
sin (2x) − −e−2x sin (2x)dx
2
2
Z
1 −2x
=− e
cos (2x) + sin (2x) − e−2x sin (2x)dx
2
Z
1
e−2x sin (2x)dx = − e−2x cos (2x) + sin (2x) + C
2
Z
1
∴ e−2x sin (2x)dx = − e−2x cos (2x) + sin (2x) + C
4
2
2
MAT 21B: Winter 2017
Homework
Due February 17
1 : u1 = e−2x , du1 = −2e−2x , v1 = − 12 cos (2x), and dv1 = sin (2x)
2 : u2 = e−2x , du2 = −2e−2x , v2 = 12 sin (2x), and dv2 = cos (2x)
8.2.32 skip
R 0.5π
8.2.48 0 x3 cos (2x)dx
Following the example 8 on page 466,
3
0.5π
Z 0.5π
x
3x2
6x
6
3
x cos (2x)dx =
sin (2x) +
cos (2x) −
sin (2x) −
cos (2x)
2
4
8
16
0
0
3π 2
3
3
= 0−
−0+
− 0+0−0−
16
8
8
2
3
3π
+
=−
16
4
8.2.53.a) skip
8.2.69 Show that
Let g(x) =
Rb
x
R b R b
a
x
Rb
f (t)dt dx = a (x − a)f (x)dx
Rx
f (t)dt = − b f (t)dt. Then, g 0 (x) = −f (x) by the Fundamental Theorem of Calculus.
Z b
Z b Z b
g(x)dx
f (t)dt dx =
a
a
x
b Z b
1
xg 0 (x)dx
= xg(x) −
a
a
Z
b
xf (x)dx
= b · g(b) − a · g(a) +
a
Z b
Z b
Z b
=b
f (x)dx − a
f (x)dx +
xf (x)dx
b
a
a
Z b
=0+
− af (x) + xf (x) dx
a
Z b
=
(x − a)f (x)dx
a
0
1 : u = g(x), du = g (x), v = x, and dv = 1
8.3.1 skip
8.3.5
R
sin3 (x)dx
Z
Z
Z 3
2
sin (x)dx = sin (x) sin (x)dx =
1 − cos2 (x) sin (x)dx
Z 1
=
sin (x) − cos2 (x) sin (x) dx = − cos (x) + cos3 (x) + C
3
3
MAT 21B: Winter 2017
Homework
8.3.22 skip
Hint: change cos2 (2θ) to 1 − sin2 (2θ)
8.3.35
R
sec3 (x) tan (x)dx
Z
Z
1
3
2
sec (x) tan (x)dx = sec (x) sec (x) tan (x) dx = sec3 (x) + C
3
8.3.38
R
sec4 (x) tan2 (x)dx
Z
Z
4
2
sec (x) tan (x)dx = sec2 (x) sec2 (x) tan2 (x)dx
Z
2
2
= sec (x) 1 + tan (x) tan2 (x)dx
Z =
sec2 (x) tan2 (x) + sec2 (x) tan4 (x) dx
=
8.3.51
R
1
1
tan3 (x) + tan5 (x) + C
3
5
sin (3x) cos (2x)dx
sin (A
± B) = sin (A) cos (B) ±cos (A) sin (B)
∴ 21 sin (A + B) + sin (A − B) = sin (A) cos (B)
Z
Z
1
sin (3x + 2x) + sin (3x − 2x) dx
2
Z 1
=
sin (5x) + sin (x) dx
2
1
1
= − cos (5x) − cos (x) + C
10
2
sin (3x) cos (2x)dx =
4
Due February 17