1000 Solved Problems in Modern Physics

(Grace) #1

52 1 Mathematical Physics


=−

1

5

ln(2x+1)+

1

5

ln(x−2)+C

=

1

5

ln

(

x− 2
2 x+ 1

)

+C

1.42r^2 =a^2 sin 2θ
Elementary area


dA=

1

2

r^2 dθ

A=

1

2

∫π/ 2

0

r^2 dθ=

a^2
2

∫π/ 2

0

sin 2θdθ=

a^2

∫π/ 2

0

sinθd(sinθ)=

a^2
2

sin^2 θ|^10 =

a^2
2

Fig. 1.11Polar diagram of
the curver^2 =a^2 sin 2θ


1.43 Sincex^2 +2 occurs twice as a factor, assume


x^3 +x^2 + 2
(x^2 +2)^2

=

Ax+B
(x^2 +2)^2

+

Cx+D
x^2 + 2
On clearing off the fractions, we get
x^3 +x^2 + 2 =Ax+B+(Cx+D)(x^2 +2)
orx^3 +x^2 + 2 =Cx^3 +Dx^2 +(A+ 2 C)x+B+ 2 D
Equating the coefficients of like powers ofx
C= 1 ,D= 1 ,A+ 2 C= 0 ,B+ 2 D= 2
This givesA=− 2 ,B= 0 ,C= 1 ,D= 1
Hence,
x^3 +x^2 + 2
(x^2 +2)^2

=−

2 x
(x^2 +2)^2

+

x
x^2 + 2

+

1

x^2 + 2

(x^3 +x^2 +2)dx
(x^2 +2)^2

=−


2 xdx
(x^2 +2)^2

+


xdx
x^2 + 2

+


dx
x^2 + 2

=

1

x^2 + 2

+

1

2

ln(x^2 +2)+

1


2

tan−^1

(

x

2

)

+C

1.44

∫∞

0

4 a^3 dx
x^2 + 4 a^2

=lim
b=∞

∫b

0

4 a^3 dx
x^2 + 4 a^2

=lim
b=∞

[

2 a^2 tan−^1

(x

2 a

)]b

0

=limb=∞

[

2 a^2 tan−^1

(

b
2 a

)]

= 2 a^2.

π
2

=πa^2
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