Question

$\int_{–\pi}^{\pi}\frac{2x(1 + \sin x)}{1 + \cos^{2}x}$dx is –

• A. p²/4
• B. p²
• C. Zero
• D. p/2

Answer 1

Answer: (B)

p²

Answer 2

$\int_{–\pi}^{\pi}\frac{2x(1 + \sin x)}{1 + \cos^{2}x}$dx

= $\int _ { - \pi } ^ { \pi } \frac { 2 x } { 1 + \cos ^ { 2 } x } d x + 2$ $\int_{–\pi}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}$dx

= 0 + 4 $\int_{0}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}$dx Ž 4 $\int_{0}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}$ dx…(1)

Now I =$\int_{0}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}$dx =$\int_{0}^{\pi}\frac{(\pi - x)\sin(\pi - x)}{1 + \cos^{2}(\pi - x)}$dx

= $\int_{0}^{\pi}\frac{(\pi - x)\sin x}{1 + \cos^{2}x}$ dx

I = dx – $\int_{0}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}$dx

Ž 2I = p$\int_{0}^{\pi}\frac{\sin x}{1 + \cos^{2}x}$dx

Ž $\frac { \pi } { 2 }$ $\int_{0}^{\pi}\frac{\sin x}{1 + \cos^{2}x}$dx

\ From (1),

$\int _ { - \pi } ^ { \pi } \frac { 2 x ( 1 + \sin x ) } { 1 + \cos ^ { 2 } x }$ dx = 4$\left( \frac { \pi } { 2 } \right)$ $\int_{0}^{\pi}\frac{\sin x}{1 + \cos^{2}x}$dx

= 2p $\int_{0}^{\pi}\frac{\sin x}{1 + \cos^{2}x}$dx

Put cos x = t so that – sin x dx = dt

i.e. sinx dx = – dt.

When x = 0, t = 1 ; when x = p, t = –1

\ $\int_{- \pi}^{\pi}\frac{2x(1 + \sin x)}{1 + \cos^{2}x}$dx = 2p $\int_{1}^{–1}\frac{- dt}{1 + t^{2}}$

= –2p $\left\lbrack \tan^{- 1}t \right\rbrack_{}^{–1}$

= –2p [tan^–1(–1) – tan^–1 (1)]

= –2p$\left\lbrack –\frac{\pi}{4} - \frac{\pi}{4} \right\rbrack$ = p².