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$\int \frac{\operatorname{cosec} x}{\cos ^{2}\left(1+\log \tan \frac{x}{2}\right)} d x=$
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Verified Answer
The correct answer is:
$\tan [1+\log \tan x / 2]+c$
Let $I=\int \frac{\operatorname{cosec} x}{\cos ^{2}\left(1+\log \tan \begin{array}{l}x \\ 2\end{array}\right)} d x$
Put $1+\log \tan \frac{x}{2}=t$
$\Rightarrow \quad-\frac{1}{\tan \frac{x}{2}} \cdot \sec ^{2} \frac{x}{2} \cdot \frac{d x}{2}=d t$
$\Rightarrow \quad \frac{d x}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=d t \Rightarrow \frac{d x}{\sin x}=d t$
$\begin{aligned} \Rightarrow & \operatorname{cosec} x d x=d t \\ \text { So, } I &=\int \frac{d i}{\cos ^{2} t}=\int \sec ^{2} t d t=\tan t+c \\ &=\tan \left(1+\log \tan \frac{x}{2}\right)+i \end{aligned}$
Put $1+\log \tan \frac{x}{2}=t$
$\Rightarrow \quad-\frac{1}{\tan \frac{x}{2}} \cdot \sec ^{2} \frac{x}{2} \cdot \frac{d x}{2}=d t$
$\Rightarrow \quad \frac{d x}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=d t \Rightarrow \frac{d x}{\sin x}=d t$
$\begin{aligned} \Rightarrow & \operatorname{cosec} x d x=d t \\ \text { So, } I &=\int \frac{d i}{\cos ^{2} t}=\int \sec ^{2} t d t=\tan t+c \\ &=\tan \left(1+\log \tan \frac{x}{2}\right)+i \end{aligned}$
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