int sqrt{operatorname{cosec} x - sin x} , dx | vedclass

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Question

(A) Let $I = \int \sqrt{\operatorname{cosec} x - \sin x} \, dx$ $I = \int \sqrt{\frac{1}{\sin x} - \sin x} \, dx = \int \sqrt{\frac{1 - \sin^2 x}{\sin

Vedclass · Accepted Answer

$2 \sqrt{\sin x} + C$

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(A) Let $I = \int \sqrt{\operatorname{cosec} x - \sin x} \, dx$ $I = \int \sqrt{\frac{1}{\sin x} - \sin x} \, dx = \int \sqrt{\frac{1 - \sin^2 x}{\sin x}} \, dx$ $I = \int \frac{\sqrt{\cos^2 x}}{\sqrt{\sin x}} \, dx = \int \frac{\cos x}{\sqrt{\sin x}} \, dx$ Let $u = \sin x$,then $du = \cos x \, dx$ $I = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} \, du$ $I = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$ Substituting $u = \sin x$,we get $I = 2\sqrt{\sin x} + C$

$\int \sqrt{\operatorname{cosec} x - \sin x} \, dx$ is equal to

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