Evaluate the integral int frac{sqrt{cot x}}{s | vedclass

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Question

(C) Let $I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sqrt{\cot x}}{\frac{\

Vedclass · Accepted Answer

$-2 \sqrt{\cot x}$

Vedclass · Answer

(C) Let $I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sqrt{\cot x}}{\frac{\sin x \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} d x = \int \frac{\sqrt{\cot x}}{	an x} \sec^2 x d x$. Since $\frac{1}{	an x} = \cot x$,we have $I = \int \sqrt{\cot x} \cdot \cot x \cdot \sec^2 x d x = \int (\cot x)^{3/2} \sec^2 x d x$. Alternatively,rewrite the denominator: $I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = \int \frac{\sqrt{\cot x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\cot x \cdot \cos^2 x} d x = \int \frac{\sec^2 x}{\sqrt{\cot x}} d x$. Let $u = \cot x$,then $du = -\csc^2 x d x$,so $d x = -\frac{du}{\csc^2 x} = -\sin^2 x du$. $I = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{\sin x \cos x} \cdot (-\sin^2 x du) = -\int \frac{\sin x}{\cos x \sqrt{u}} du = -\int \frac{1}{u \sqrt{u}} du = -\int u^{-3/2} du$. $I = -\left( \frac{u^{-1/2}}{-1/2} ight) + c = 2 \sqrt{u} + c = 2 \sqrt{\cot x} + c$. Given $\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = -f(x) + c$,we have $-f(x) = 2 \sqrt{\cot x}$,so $f(x) = -2 \sqrt{\cot x}$.

Evaluate the integral $\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = -f(x) + c$. Find $f(x)$.

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