int frac{operatorname{cosec} x , dx}{cos ^2le | vedclass

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Question

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log 	an \frac{x}{2}ight)}$.Let $t = 1+\log \left(	an \frac{x}{2}ight)$.Dif

Vedclass · Accepted Answer

$	an \left(1+\log 	an \frac{x}{2}ight)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log 	an \frac{x}{2}ight)}$.Let $t = 1+\log \left(	an \frac{x}{2}ight)$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{	an \frac{x}{2}} \cdot \sec ^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos ^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \operatorname{cosec} x$.Thus,$\operatorname{cosec} x \, dx = dt$.Substituting these into the integral,we get:$I = \int \frac{1}{\cos ^2 t} \, dt = \int \sec ^2 t \, dt$.Integrating,we get $I = 	an t + c$.Substituting back $t = 1+\log \left(	an \frac{x}{2}ight)$,we get $I = 	an \left(1+\log 	an \frac{x}{2}ight)+c$.

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

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