Question

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

• A. $- tan [1 + Log \; Tan \frac{x}{2}]+c$
• B. $Sec^2 [1+ Log \; Tan \frac{x}{2}]+c$
• C. $tan [1 + Log \; Tan \frac{x}{2}]+c$
• D. $Sin^2 [1 + Log \; Tan \frac{x}{2}]+c$

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}\; x dx =dt $
$ \therefore I = \int \frac{dt}{\cos^{2}t } = \int\sec^{2} t dt$
$ = \tan t +c $
$= \tan \left(1+ \log \tan \frac{x}{2}\right) +c $