The value of int(1-cos x) operatorname{cosec} | vedclass

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(B) Let $I = \int (1 - \cos x) \operatorname{cosec}^2 x \, dx$ Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2}

Vedclass · Accepted Answer

$	an \frac{x}{2} + c$,where $c$ is a constant of integration.

Vedclass · Answer

(B) Let $I = \int (1 - \cos x) \operatorname{cosec}^2 x \, dx$ Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have: $I = \int \frac{2 \sin^2 \frac{x}{2}}{\sin^2 x} \, dx$ $I = \int \frac{2 \sin^2 \frac{x}{2}}{(2 \sin \frac{x}{2} \cos \frac{x}{2})^2} \, dx$ $I = \int \frac{2 \sin^2 \frac{x}{2}}{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}} \, dx$ $I = \frac{1}{2} \int \frac{1}{\cos^2 \frac{x}{2}} \, dx$ $I = \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$ Integrating $\sec^2 \frac{x}{2}$,we get: $I = \frac{1}{2} \cdot \frac{	an \frac{x}{2}}{1/2} + c$ $I = 	an \frac{x}{2} + c$

The value of $\int(1-\cos x) \operatorname{cosec}^2 x \, dx$ is

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