Question

$\int \frac{\text{cosecx }}{{\cos }^2(1+LogTan\frac{x}{2})}dx$=

• A. $-tan[1+LogTan\frac{x}{2}]+c$
• B. $Se{c}^2[1+LogTan\frac{x}{2}]+c$
• C. $tan[1+LogTan\frac{x}{2}]+c$
• D. $Si{n}^2[1+LogTan\frac{x}{2}]+c$

Answer 1

Answer: (C)

Let $I=\int \frac{\text{cosec }x}{{\cos }^2\left(1+\log \tan \frac{x}{2}\right)}dx$
Put $1+\log \tan \frac{x}{2}=t$
$\Rightarrow \frac{1}{\tan \frac{x}{2}}.{\sec }^2\frac{x}{2}.\frac{1}{2}dx=dt$
$\Rightarrow \frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}dx=dt$
$\Rightarrow \text{cosec}xdx=dt$
$\therefore I=\int \frac{dt}{{\cos }^{2}t}=\int {\sec }^{2}tdt$
$=\tan t+c$
$=\tan \left(1+\log \tan \frac{x}{2}\right)+c$