int frac{sqrt{cot x}}{sin 2x} dx = | vedclass

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Question

(B) Let $I = \int \frac{\sqrt{\cot x}}{\sin 2x} dx$. Using the identity $\sin 2x = 2 \sin x \cos x$, we have: $I = \int \frac{\sqrt{\cot x}}{2 \sin x

Vedclass · Accepted Answer

$-\sqrt{\cot x} + C$

Vedclass · Answer

(B) Let $I = \int \frac{\sqrt{\cot x}}{\sin 2x} dx$. Using the identity $\sin 2x = 2 \sin x \cos x$, we have: $I = \int \frac{\sqrt{\cot x}}{2 \sin x \cos x} dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sqrt{\cot x}}{2 	an x \cos^2 x} dx = \frac{1}{2} \int \frac{\sqrt{\cot x}}{	an x} \sec^2 x dx$. Since $\frac{1}{	an x} = \cot x$, we get: $I = \frac{1}{2} \int \sqrt{\cot x} \cdot \cot x \cdot \sec^2 x dx = \frac{1}{2} \int (\cot x)^{3/2} \sec^2 x dx$. Wait, let us re-evaluate: $I = \int \frac{\sqrt{\cot x}}{2 \sin x \cos x} dx = \int \frac{\sqrt{\cot x}}{2 \sin^2 x \cot x} dx = \frac{1}{2} \int \frac{\operatorname{cosec}^2 x}{\sqrt{\cot x}} dx$. Let $t = \cot x$, then $dt = -\operatorname{cosec}^2 x dx$, so $\operatorname{cosec}^2 x dx = -dt$. $I = \frac{1}{2} \int \frac{-dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-1/2} dt$. $I = -\frac{1}{2} \cdot \frac{t^{1/2}}{1/2} + C = -\sqrt{t} + C$. Substituting $t = \cot x$, we get $I = -\sqrt{\cot x} + C$.

$\int \frac{\sqrt{\cot x}}{\sin 2x} dx =$

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