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$$
\int \frac{\operatorname{cosec} x \mathrm{~d} x}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)}=
$$
Options:
\int \frac{\operatorname{cosec} x \mathrm{~d} x}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)}=
$$
Solution:
2607 Upvotes
Verified Answer
The correct answer is:
$\tan \left(1+\log \left(\tan \frac{x}{2}\right)\right)+\mathrm{c}$, where $\mathrm{c}$ is constant of integration
Let $\mathrm{I}=\int \frac{\operatorname{cosec} x \mathrm{~d} x}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)} \mathrm{d} x$
Let $1+\log \left(\tan \frac{x}{2}\right)=\mathrm{t}$
Differentiating both sides w.r.t. t, we get
$$
\begin{aligned}
& \frac{1}{\tan \frac{x}{2}} \sec ^2 \frac{x}{2} \times \frac{1}{2} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \operatorname{cosec} x \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \mathrm{I}=\int \frac{1}{\cos ^2 \mathrm{t}} \mathrm{dt}=\int \sec ^2 \mathrm{t} d \mathrm{t} \\
& =\tan (\mathrm{t})+\mathrm{c} \\
& =\tan \left(1+\log \left(\tan \frac{x}{2}\right)\right)+c \\
&
\end{aligned}
$$
Let $1+\log \left(\tan \frac{x}{2}\right)=\mathrm{t}$
Differentiating both sides w.r.t. t, we get
$$
\begin{aligned}
& \frac{1}{\tan \frac{x}{2}} \sec ^2 \frac{x}{2} \times \frac{1}{2} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \operatorname{cosec} x \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad \mathrm{I}=\int \frac{1}{\cos ^2 \mathrm{t}} \mathrm{dt}=\int \sec ^2 \mathrm{t} d \mathrm{t} \\
& =\tan (\mathrm{t})+\mathrm{c} \\
& =\tan \left(1+\log \left(\tan \frac{x}{2}\right)\right)+c \\
&
\end{aligned}
$$
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