int sqrt{1+2 cot x(cot x+operatorname{cosec} | vedclass

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Question

(A) Let $I = \int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx$. Since $1 = \operatorname{cosec}^2 x - \cot^2 x$,we have: $I = \int \sqrt{\o

Vedclass · Accepted Answer

$2 \log \left(\sin \frac{x}{2}\right)+C$

Vedclass · Answer

(A) Let $I = \int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx$. Since $1 = \operatorname{cosec}^2 x - \cot^2 x$,we have: $I = \int \sqrt{\operatorname{cosec}^2 x - \cot^2 x + 2 \cot^2 x + 2 \cot x \operatorname{cosec} x} \, dx$ $I = \int \sqrt{\operatorname{cosec}^2 x + \cot^2 x + 2 \cot x \operatorname{cosec} x} \, dx$ $I = \int \sqrt{(\operatorname{cosec} x + \cot x)^2} \, dx$ $I = \int (\operatorname{cosec} x + \cot x) \, dx$ Using the standard integrals $\int \operatorname{cosec} x \, dx = \log |\operatorname{cosec} x - \cot x|$ and $\int \cot x \, dx = \log |\sin x|$: $I = \log |\operatorname{cosec} x - \cot x| + \log |\sin x| + C$ $I = \log \left| \frac{1 - \cos x}{\sin x} \cdot \sin x \right| + C$ $I = \log |1 - \cos x| + C$ Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$: $I = \log |2 \sin^2 \frac{x}{2}| + C$ $I = \log 2 + 2 \log |\sin \frac{x}{2}| + C$ Since $\log 2$ is a constant,we can absorb it into $C$: $I = 2 \log |\sin \frac{x}{2}| + C$

$\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx = $

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