$\int \frac{\operatorname{cosec} x d x}{\cos ^2\left(1+\log an \frac{x}{2} ight)}=$

Question

$\int \frac{\operatorname{cosec} x d x}{\cos ^2\left(1+\log 	an \frac{x}{2}ight)}=$

Aicharya · Accepted Answer

Let $\mathrm{I}=\int \frac{\operatorname{cosec} x \mathrm{~d} x}{\cos ^2\left(1+\log 	an \frac{x}{2}ight)} \mathrm{d} x$ Let $1+\log \left(	an \frac{x}{2}ight)=\mathrm{t}$ Differentiating both sides w.r.t. t, we get $\frac{1}{	an \frac{x}{2}} \sec ^2 \frac{x}{2} 	imes \frac{1}{2} \mathrm{~d} x=\mathrm{dt}$ $\begin{aligned} & 	herefore \quad \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \mathrm{~d} x=\mathrm{dt} \ & 	herefore \quad \operatorname{cosec} x \mathrm{~d} x=\mathrm{dt} \ & 	herefore \quad I=\int \frac{1}{\cos ^2 t} d t \ & =\int \sec ^2 t d t \ & =	an (\mathrm{t})+\mathrm{c} \ & =	an \left(1+\log \left(	an \frac{x}{2}ight)ight)+\mathrm{c} \end{aligned}$

$\int \frac{\operatorname{cosec} x d x}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)}=$

Solution