Q. $\int_{}^{}{cosec\ (x - a)cosec\ x\ dx}$ is equal to

A) $\frac{- 1}{\sin a}\log{\left| \sin x\ cosec(x - a) \right| + c}$
B) $\frac{- 1}{\sin a}\log{\lbrack\sin{(x - a)\sin{x\rbrack + c}}}$
C) $\frac{1}{\sin a}\log{\lbrack\sin{(x - a)cosec\ x\rbrack + c}}$
D) $\frac{1}{\sin a}\log{\lbrack\sin{(x - a)\sin{x\rbrack + c}}}$
Explanation

(a)
Let $I = \int_{}^{}{cosec\ (x - a)cosec\ x\ dx}$
$I = \int_{}^{}{\frac{\sin a}{\sin a\sin{(x - a)}\sin x}\ dx}$
$= - \frac{1}{\sin a}\int_{}^{}\left\lbrack \frac{\sin{(x - a)\cos{x - \cos{(x - a)\sin x}}}}{\sin{(x - a)\sin x}} \right\rbrack dx$
$= - \frac{1}{\sin a}\int_{}^{}\left\lbrack \cot{x - \cot\ (x - a)} \right\rbrack dx$
$= - \frac{1}{\sin a}\left\lbrack \log{|\sin{x|}} - \log\left| \sin{(x - a)} \right| \right\rbrack + c$
$= - \frac{1}{\sin a}\lbrack\log{\left| \sin x\ cosec(x - a) \right|\rbrack + c}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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