If int(1-cos x) operatorname{cosec}^2 x , dx | vedclass

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Question

(A) We are given the integral $I = \int(1-\cos x) \operatorname{cosec}^2 x \, dx$. Using the trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}

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$	an \frac{x}{2}$

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(A) We are given the integral $I = \int(1-\cos x) \operatorname{cosec}^2 x \, dx$. Using the trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have: $I = \int (2 \sin^2 \frac{x}{2}) \cdot \frac{1}{\sin^2 x} \, dx$ $I = \int \frac{2 \sin^2 \frac{x}{2}}{(2 \sin \frac{x}{2} \cos \frac{x}{2})^2} \, dx$ $I = \int \frac{2 \sin^2 \frac{x}{2}}{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}} \, dx$ $I = \int \frac{1}{2 \cos^2 \frac{x}{2}} \, dx$ $I = \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$ Integrating $\sec^2 \frac{x}{2}$,we get: $I = \frac{1}{2} \cdot \frac{	an \frac{x}{2}}{1/2} + c = 	an \frac{x}{2} + c$. Comparing this with $f(x) + c$,we find $f(x) = 	an \frac{x}{2}$.

If $\int(1-\cos x) \operatorname{cosec}^2 x \, dx = f(x) + c$,then $f(x)$ is equal to

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