int (cos x) log cot (frac{x}{2}) dx = | vedclass

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(C) Using integration by parts,let $u = \log \cot (\frac{x}{2})$ and $dv = \cos x dx$. Then $du = \frac{1}{\cot (\frac{x}{2})} \cdot (-\csc^2 (\frac{x

Vedclass · Accepted Answer

$(\sin x) \log \cot (\frac{x}{2}) + x + c$

Vedclass · Answer

(C) Using integration by parts,let $u = \log \cot (\frac{x}{2})$ and $dv = \cos x dx$. Then $du = \frac{1}{\cot (\frac{x}{2})} \cdot (-\csc^2 (\frac{x}{2})) \cdot \frac{1}{2} dx = -\frac{1}{2 \sin (\frac{x}{2}) \cos (\frac{x}{2})} dx = -\frac{1}{\sin x} dx$. And $v = \int \cos x dx = \sin x$. Using the formula $\int u dv = uv - \int v du$: $\int \cos x \log \cot (\frac{x}{2}) dx = \sin x \log \cot (\frac{x}{2}) - \int \sin x \cdot (-\frac{1}{\sin x}) dx$ $= \sin x \log \cot (\frac{x}{2}) + \int 1 dx$ $= \sin x \log \cot (\frac{x}{2}) + x + c$.

$\int (\cos x) \log \cot (\frac{x}{2}) dx =$

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