Question

$\int cosec(x-a)cosecxdx$ is equal to

• A. $\frac{1}{\sin a}\log |\sin (x-a)\sin x|+C$
• B. $\frac{-1}{\sin a}\log |\sin xcosec(x-a)|+C$
• C. $\frac{-1}{\sin a}\log |\sin (x-a)\sin x|+C$
• D. $-\frac{1}{\sin a}\log |\sin (x-a)\mathrm{cosec}x|+C$

Answer 1

Answer: (B)

Let $l=\int \mathrm{cosec}(x-a)\mathrm{cosec}xdx$
$=\int \frac{\sin a}{\sin a\sin (x-a)\sin x}dx$
$=-\frac{1}{\sin a}\int \frac{\sin [(x-a)-x]}{\sin (x-a)\sin }dx$
$=-\frac{1}{\sin a}\int \left[\frac{\sin (x-a)\cos x-\cos (x-a)\sin x}{\sin (x-a)\sin x}\right]dx$
$=-\frac{1}{\sin a}\int [\cot x-\cot (x-a]dx$
$=-\frac{1}{\sin a}[\log |\sin x|-\log |\sin (x-a)|]+C$
$=-\frac{1}{\sin a}[\log ||\sin x\mathrm{cosec}(x-a)\mid ]+C$