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

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

Vedclass · Accepted Answer

$	an \left(\frac{x}{2}ight)+C$

Vedclass · Answer

(A) Let $I = \int (1 - \cos x) \operatorname{cosec}^2 x \, dx$. Using the identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}ight)$ and $\operatorname{cosec}^2 x = \frac{1}{\sin^2 x}$,we have: $I = \int 2 \sin^2 \left(\frac{x}{2}ight) \cdot \frac{1}{\sin^2 x} \, dx$. Since $\sin x = 2 \sin \left(\frac{x}{2}ight) \cos \left(\frac{x}{2}ight)$,then $\sin^2 x = 4 \sin^2 \left(\frac{x}{2}ight) \cos^2 \left(\frac{x}{2}ight)$. Substituting this into the integral: $I = \int \frac{2 \sin^2 \left(\frac{x}{2}ight)}{4 \sin^2 \left(\frac{x}{2}ight) \cos^2 \left(\frac{x}{2}ight)} \, dx$. $I = \int \frac{1}{2 \cos^2 \left(\frac{x}{2}ight)} \, dx = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}ight) \, dx$. Integrating $\sec^2 \left(\frac{x}{2}ight)$,we get $2 	an \left(\frac{x}{2}ight)$. $I = \frac{1}{2} \cdot 2 	an \left(\frac{x}{2}ight) + C = 	an \left(\frac{x}{2}ight) + C$.

$\int(1-\cos x) \operatorname{cosec}^2 x \, dx$ is equal to

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