Q. $\int_{}^{}\frac{cosec\ x}{\cos^{2}\left( 1 + \log{\tan\frac{x}{2}} \right)}\ dx$ is equal to
A) $\sin^{2}{\left\lbrack 1 + \log{\tan\frac{x}{2}} \right\rbrack + c}$
B) $\tan\left\lbrack 1 + \log{\tan\frac{x}{2}} \right\rbrack + c$
C) $\sec^{2}\left\lbrack 1 + \log{\tan\frac{x}{2}} \right\rbrack + c$
D) $- \tan\left\lbrack 1 + \log{\tan\frac{x}{2}} \right\rbrack + c$
Explanation
(b)
Let $I = \int_{}^{}\frac{cosec\ x}{\cos^{2}{(1 + \log{\tan{\frac{x}{2})}\ }}}dx$
Put $1 + \log{\tan\frac{x}{2}} = t$
$\Rightarrow \frac{1}{\tan\frac{x}{2}}.\sec^{2}\frac{x}{2}.\frac{1}{2}dx = dt$
$\Rightarrow cosec\ x\ dx = dt$
$\therefore I = \int_{}^{}\frac{dt}{\cos^{2}t} = \int_{}^{}{\sec^{2}t}\ dx = \tan t + c$
$= \tan\left( 1 + \log{\tan\frac{x}{2}} \right) + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.