Q.18.\int\frac{\mathrm{cosec}x \,dx}{\cos^2\left(1+\log\tan\frac{x}{2}\right)}=
A. \tan\left(1+\log\left(\tan\frac{x}{2}\right)\right)+C, where C is the constant of integration.
B. \tan(1+\log(\tan x))+C, where C is the constant of integration.
C. \tan\left(\log\left(\tan\frac{x}{2}\right)\right)+C, where C is the constant of integration.
D. \tan\left(\tan\frac{x}{2}\right)+C, where C is the constant of integration.
Answer: A
Explanation :-
I=\int{\frac{\operatorname{cosec}x\,dx}{\cos^{2}\left(1+\log\tan{\frac{x}{2}}\right)}}Let 1+\log\left(\tan{\frac{x}{2}}\right)=t
Differentiating both sides w.r.t. x, we get:
\frac{1}{\tan{\frac{x}{2}}} \sec^{2}{\frac{x}{2}} \times \frac{1}{2} \,dx = dt \frac{1}{2 sin \frac{x}{2}cos\frac{x}{2}}dx=dtcosec x dx = dt
I=\int \frac{1}{cos^{2}t}dt=\int sec^{2}t\; dt= tan (t) + c
= tan \left ( 1+log\left ( tan\frac{x}{2} \right ) \right )+c