Indefinite Integration 1 Question 82
83. Evaluate $\int _0^{\pi} \frac{x d x}{1+\cos \alpha \sin x}, 0<\alpha<\pi$.
$\left(1986,2 \frac{1}{2} M\right)$
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Answer:
Correct Answer: 83. $\frac{\alpha \pi}{\sin \alpha}$
Solution:
- Let $I=\int _0^{\pi} \frac{x}{1+\cos \alpha \sin x} d x$
$\Rightarrow \quad I=\int _0^{\pi} \frac{(\pi-x)}{1+\cos \alpha \sin (\pi-x)} d x$
$\Rightarrow \quad I=\int _0^{\pi} \frac{(\pi-x)}{1+\cos \alpha \sin x} d x$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} & 2 I=\pi \int _0^{\pi} \frac{d x}{1+\cos \alpha \sin x} \\ & \Rightarrow \quad 2 I=\pi \int _0^{\pi} \frac{\sec ^{2} \frac{x}{2} d x}{\left(1+\tan ^{2} \frac{x}{2}\right)+2 \cos \alpha \tan \frac{x}{2}} \end{aligned} $$
Put $\tan \frac{x}{2}=t \Rightarrow \sec ^{2} \frac{x}{2} d x=2 d t$
$\therefore \quad 2 I=\pi \int _0^{\infty} \frac{2 d t}{1+t^{2}+2 t \cos \alpha}$
$\Rightarrow \quad 2 I=2 \pi \int _0^{\infty} \frac{d t}{(t+\cos \alpha)^{2}+\sin ^{2} \alpha}$
$$ \begin{aligned} I & =\frac{\pi}{\sin \alpha} \tan ^{-1} \frac{t+\cos \alpha}{\sin \alpha}{ } _0^{\infty} \\ & =\frac{\pi}{\sin \alpha}\left[\tan ^{-1}(\infty)-\tan ^{-1}(\cot \alpha)\right] \\ & =\frac{\pi}{\sin \alpha} \frac{\pi}{2}-\frac{\pi}{2}-\alpha=\frac{\alpha \pi}{\sin \alpha} \\ \therefore \quad I & =\frac{\alpha \pi}{\sin \alpha} \end{aligned} $$
