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

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Question

(A) Given integral $I = \int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx$. Using the identity $1+\cot^2 x = \operatorname{cosec}^2 x$, we e

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$2 \log \left|\sin \frac{x}{2}\right|+c$

Vedclass · Answer

(A) Given integral $I = \int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx$. Using the identity $1+\cot^2 x = \operatorname{cosec}^2 x$, we expand the expression inside the square root: $I = \int \sqrt{1+2 \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$. Since $1+\cot^2 x = \operatorname{cosec}^2 x$, we have $1 = \operatorname{cosec}^2 x - \cot^2 x$. Substituting this: $I = \int \sqrt{\operatorname{cosec}^2 x - \cot^2 x + 2 \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$ $I = \int \sqrt{\operatorname{cosec}^2 x + \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$ $I = \int \sqrt{(\operatorname{cosec} x + \cot x)^2} \, dx$ $I = \int (\operatorname{cosec} x + \cot x) \, dx$ Integrating term by term: $I = \ln |\operatorname{cosec} x - \cot x| + \ln |\sin x| + c$ $I = \ln |(\operatorname{cosec} x - \cot x) \sin x| + c$ $I = \ln |1 - \cos x| + c$ Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$: $I = \ln |2 \sin^2 \frac{x}{2}| + c = \ln 2 + 2 \ln |\sin \frac{x}{2}| + c$ Since $\ln 2$ is a constant, we can absorb it into $c$: $I = 2 \ln |\sin \frac{x}{2}| + c$.

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

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