Q90 / 180 · KCET 2009 · Mathematics
$\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x$ is equal to
- Correct answerA$\frac{-1}{\sin a} \log |\sin x \operatorname{cosec}(x-a)|+c$
- B$\frac{-1}{\sin a} \log [\sin (x-a) \sin x]+c$
- C$\frac{1}{\sin a} \log [\sin (x-a) \operatorname{cosec} x]+c$
- D$\frac{1}{\sin a} \log [\sin (x-a) \sin x]+c$
Solution
Let $I=\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x$
$=\int \frac{\sin a}{\sin a \sin (x-a) \sin x} d x$
$=-\frac{1}{\sin a} \int \frac{\sin [(x-a)-x]}{\sin (x-a) \sin x} d x$
$=-\frac{1}{\sin a} \int\left[\frac{\sin (x-a) \cos x-\cos (x-a) \sin x}{\sin (x-a) \sin x}\right] d x$
$=-\frac{1}{\sin a} \int[\cot x-\cot (x-a)] d x$
$=-\frac{1}{\sin a}[\log |\sin x|-\log |\sin (x-a)|]+c$
$=\frac{-1}{\sin a}[\log |\sin x \operatorname{cosec}(x-a)|]+c$
Hence, the correct answer is option: 'A'.