Evaluate int frac{sqrt{cot x}}{sin x cos x} d | vedclass

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Question

(D) 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

(D) 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 integral as: $I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = \int \frac{\sqrt{\cot x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{	an x \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\cot x} \cdot \frac{1}{\sin^2 x} d x$ is not correct. Let $t = \cot x$, then $dt = -\csc^2 x d x$, so $dx = -\sin^2 x dt$. $I = \int \frac{\sqrt{t}}{\sin x \cos x} (- \sin^2 x) dt = \int \frac{\sqrt{t}}{\cos x / \sin x} (-dt) = \int \frac{\sqrt{t}}{t} (-dt) = -\int t^{-1/2} dt$. $I = -[2 t^{1/2}] + c = -2 \sqrt{\cot x} + c$. Comparing with $-f(x) + c$, we get $-f(x) = -2 \sqrt{\cot x}$, so $f(x) = 2 \sqrt{\cot x}$.

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

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