1.3 Solutions 51
1.38 f(a+x)=sin(a+x)
Putx= 0
f(a)=sina
f′(a)=cosa
f′′(a)=−sina
f′′′(a)=−cosa
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)=sina+
x
1
cosa−
x^2
2!
sina−
x^3
3!
cosa+···
1.39 We know that
y= 1 +x+x^2 +x^3 +x^4 +··· = 1 /(1−x)
Differentiating with respect tox,
dy/dx= 1 + 2 x+ 3 x^2 + 4 x^3 +··· = 1 /(1−x)^2 =S
1.3.7 Integration........................................
1.40 (a)
∫
sin^3 xcos^6 xdx=
∫
sin^2 xcos^6 xsinxdx
=−
∫
(1−cos^2 x) cos^6 xd(cosx)=
∫
cos^8 xd(cosx)−
∫
cos^6 xd(cosx)
=
cos^9 x
9
−
cos^7 x
7
+C
(b)
∫
sin^4 xcos^2 xdx=
∫
(sinxcosx)^2 sin^2 xdx
=
∫
1
4
sin^22 x(
1
2
−
1
2
cos 2x)dx
=
1
8
∫
sin^22 xdx−
1
8
∫
sin^22 xcos 2xdx
=
1
8
∫
(
1
2
−
1
2
cos 4x)dx−
1
8
∫
sin^22 xcos 2xdx
=
x
16
−
sin 4x
64
−
sin^32 x
48
+C
1.41 Express the integrand as sum of functions.
Let
1
2 x^2 − 3 x− 2
=
1
(2x+1)(x−2)
=
A
2 x+ 1
+
B
x− 2
=
A(x−2)+B(2x+1)
(2x+1)(x−2)
B− 2 A= 1
A+ 2 B= 0
Solving,A=−^25 andB=^15
I=−
2
5
∫
dx
2 x+ 1
+
1
5
∫
dx
x− 2