1.3 Solutions
51
1.38 f (a + x) = sin(a + x)
Put x = 0
f (a) = sin a
f
(a) = cos a
f
(a) = − sin a
f
(a) = − cos a
Substitute in
f (x) = f (a) +
(x − a)
1!
f
(a) +
(x − a)
2
2!
f
(a) +
(x − a)
3
3!
f
(a) + · · ·
sin(a + x) = sin a +
x
1
cos a −
x
2
2!
sin a −
x
3
3!
cos a + · · ·
1.39 We know that
y = 1 + x + x
2
+ x
3
+ x
4
+ · · · = 1/(1 − x)
Differentiating with respect to x,
dy/dx = 1 + 2x + 3x
2
+ 4x
3
+ · · · = 1/(1 − x)
2
= S
1.3.7 Integration
1.40 (a)
sin
3 x cos
6 xdx =
sin
2 x cos
6 x sin xdx
= −
(1 − cos
2 x) cos
6 x d(cos x) =
cos
8 x d(cos x) −
cos
6 x d(cos x)
=
cos
9 x
9
−
cos
7 x
7
+ C
(b)
sin
4 x cos
2 xdx =
(sin x cos x)
2 sin
2 xdx
=
1
4
sin
2 2x(
1
2
−
1
2
cos 2x)dx
=
1
8
sin
2 2xdx −
1
8
sin
2 2x cos 2x dx
=
1
8
(
1
2
−
1
2
cos 4x)dx −
1
8
sin
2 2x cos 2xdx
=
x
16
−
sin 4x
64
−
sin
3 2x
48
+ C
1.41 Express the integrand as sum of functions.
Let
1
2x 2 − 3x − 2
=
1
(2x + 1)(x − 2)
=
A
2x + 1
+
B
x − 2
=
A(x − 2) + B(2x + 1)
(2x + 1)(x − 2)
B − 2A = 1
A + 2B = 0
Solving, A = −
2
5
and B =
1
5
I = −
2
5
dx
2x + 1
+
1
5
dx
x − 2
+ C