If value of $\int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} = \frac{A}{sin\alpha} + B (\alpha \ne 0) $ the values of A and B are

Q: If value of $\displaystyle \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} = \frac{A}{sin\alpha} + B (\alpha \ne 0) $ the values of A and B are

(a) $\displaystyle A =\frac{\pi}{2} , B = 0 $

(b) $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha} $

(c) $\displaystyle A =\frac{\pi}{6} , B = \frac{\pi}{ sin\alpha} $

(d) $\displaystyle A =\pi , B = \frac{\pi}{ sin\alpha} $

Ans: (a) , (b)

Sol: $\displaystyle I = \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} $

$\displaystyle I = \int_{0}^{\alpha} \frac{dx}{cos^2 \frac{x}{2} + sin^2 \frac{x}{2}-cos\alpha (cos^2 \frac{x}{2} – sin^2 \frac{x}{2})} $

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

$\displaystyle I = \int_{0}^{\alpha} \frac{sec^2 \frac{x}{2} dx}{2 sin^2 \frac{\alpha}{2} + 2 cos^2 \frac{\alpha}{2} tan^2 \frac{x}{2}} $

$\displaystyle I = \frac{1}{2} \int_{0}^{\alpha} \frac{sec^2 \frac{\alpha}{2} sec^2 \frac{x}{2} dx}{tan^2 \frac{\alpha}{2} + tan^2 \frac{x}{2}} $

Let $tan\frac{x}{2} = t $

$\displaystyle I = \int_{0}^{tan(\alpha/2)} \frac{sec^2 \frac{\alpha}{2} dt }{t^2 + tan^2 \frac{\alpha}{2}} $

$\displaystyle I = sec^2 \frac{\alpha}{2} cot\frac{\alpha}{2} [tan^{-1}\frac{t}{tan(\alpha/2)}]_{0}^{tan(\alpha/2)} $

$\displaystyle = \frac{2}{sin\alpha}.\frac{\pi}{4}= \frac{\pi}{2 sin\alpha}$

$\displaystyle A =\frac{\pi}{2} , B = 0 $

Also , $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha} $

Hence (a), (b) are correct answers.