int sqrt{1 + csc x} , dx equals | vedclass

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Question

(D) Let $I = \int \sqrt{1 + \frac{1}{\sin x}} \, dx = \int \sqrt{\frac{\sin x + 1}{\sin x}} \, dx$. Multiply numerator and denominator by $\sqrt{1 - \

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

Vedclass · Answer

(D) Let $I = \int \sqrt{1 + \frac{1}{\sin x}} \, dx = \int \sqrt{\frac{\sin x + 1}{\sin x}} \, dx$. Multiply numerator and denominator by $\sqrt{1 - \sin x}$: $I = \int \frac{\sqrt{(1 + \sin x)(1 - \sin x)}}{\sqrt{\sin x(1 - \sin x)}} \, dx = \int \frac{\cos x}{\sqrt{\sin x - \sin^2 x}} \, dx$. Let $t = \sin x$,then $dt = \cos x \, dx$. $I = \int \frac{dt}{\sqrt{t - t^2}} = \int \frac{dt}{\sqrt{\frac{1}{4} - (t^2 - t + \frac{1}{4})}} = \int \frac{dt}{\sqrt{(\frac{1}{2})^2 - (t - \frac{1}{2})^2}}$. Using $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + c$: $I = \sin^{-1}\left(\frac{t - 1/2}{1/2}\right) + c = \sin^{-1}(2t - 1) + c = \sin^{-1}(2 \sin x - 1) + c$. Since $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,$\sin^{-1}(2 \sin x - 1) = \frac{\pi}{2} - \cos^{-1}(2 \sin x - 1)$. Also,$\cos^{-1}(1 - 2 \sin x) = \pi - \cos^{-1}(2 \sin x - 1)$. By simplifying the trigonometric identities,the integral results in $2 \sin^{-1} \sqrt{\sin x} + c$ and $\cos^{-1}(1 - 2 \sin x) + c$ (up to a constant). Thus,the correct option is $(D)$.

$\int \sqrt{1 + \csc x} \, dx$ equals

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