The integral int sqrt{1 + 2cot x(csc x + cot | vedclass

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(A) Let $I = \int \sqrt{1 + 2\cot x \csc x + 2\cot^2 x} \,dx$.Since $1 + \cot^2 x = \csc^2 x$,we have $1 + 2\cot^2 x = \csc^2 x + \cot^2 x$.Thus,the e

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \sqrt{1 + 2\cot x \csc x + 2\cot^2 x} \,dx$.Since $1 + \cot^2 x = \csc^2 x$,we have $1 + 2\cot^2 x = \csc^2 x + \cot^2 x$.Thus,the expression inside the square root is $\csc^2 x + \cot^2 x + 2\cot x \csc x = (\csc x + \cot x)^2$.Since $0  0$,so $\sqrt{(\csc x + \cot x)^2} = \csc x + \cot x$.$I = \int (\csc x + \cot x) \,dx$.Using the standard integrals $\int \csc x \,dx = \log |\csc x - \cot x| + C$ and $\int \cot x \,dx = \log |\sin x| + C$,we get:$I = \log |\csc x - \cot x| + \log |\sin x| + C$.$I = \log \left| \frac{1 - \cos x}{\sin x} \cdot \sin x \right| + C = \log |1 - \cos x| + C$.Using $1 - \cos x = 2\sin^2 \frac{x}{2}$,we get:$I = \log |2\sin^2 \frac{x}{2}| + C = \log 2 + 2\log |\sin \frac{x}{2}| + C$.Absorbing $\log 2$ into the constant $C$,we get $I = 2\log |\sin \frac{x}{2}| + C$.

The integral $\int \sqrt{1 + 2\cot x(\csc x + \cot x)} \,dx$ for $0 < x < \frac{\pi}{2}$ is equal to (where $C$ is a constant of integration):

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