int frac{sin x cdot sec ^2 x-tan x cdot sin x | vedclass

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Question

(C) Let $I = \int \frac{\sin x \sec^2 x - 	an x \sin x + \cos x}{1 - \cos 2x} dx$. Using $1 - \cos 2x = 2 \sin^2 x$, we get: $I = \int \frac{\sin x \

Vedclass · Accepted Answer

$\frac{1}{2}\left[\sec x-\operatorname{cosec} x-\log \left|\frac{	an \left(\frac{\pi}{4}+\frac{x}{2}ight)}{	an \left(\frac{x}{2}ight)}ight|ight]+c$

Vedclass · Answer

(C) Let $I = \int \frac{\sin x \sec^2 x - 	an x \sin x + \cos x}{1 - \cos 2x} dx$. Using $1 - \cos 2x = 2 \sin^2 x$, we get: $I = \int \frac{\sin x \sec^2 x - 	an x \sin x + \cos x}{2 \sin^2 x} dx$ $I = \frac{1}{2} \int \left( \frac{\sin x \sec^2 x}{\sin^2 x} - \frac{	an x \sin x}{\sin^2 x} + \frac{\cos x}{\sin^2 x} ight) dx$ $I = \frac{1}{2} \int \left( \sec x 	an x - \sec x + \operatorname{cosec} x \cot x ight) dx$ Integrating term by term: $\int \sec x 	an x dx = \sec x$ $\int \sec x dx = \log |\sec x + 	an x| = \log |	an(\frac{\pi}{4} + \frac{x}{2})|$ $\int \operatorname{cosec} x \cot x dx = -\operatorname{cosec} x$ Thus, $I = \frac{1}{2} [\sec x - \log |	an(\frac{\pi}{4} + \frac{x}{2})| - \operatorname{cosec} x] + c$ $I = \frac{1}{2} [\sec x - \operatorname{cosec} x - \log |	an(\frac{\pi}{4} + \frac{x}{2})|] + c$ Since $	an(\frac{x}{2}) = \frac{\sin x}{1 + \cos x}$, the expression simplifies to option $(C)$.

$\int \frac{\sin x \cdot \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} d x=$

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