Question
$\int \frac{x+\sin x}{1-\cos x} d x$
✓
Answer
$\text { Let } I =\int \frac{x+\sin x}{1-\cos x} d x$
$=\int\left(\frac{x+2 \frac{\sin x}{2} \frac{\cos x}{2}}{2 \frac{\sin ^2 x}{2}}\right) d x$
$=\int\left(\frac{x}{2 \frac{\sin ^2 x}{2}}+\frac{2 \frac{\sin x}{2} \frac{\cos x}{2}}{2 \frac{\sin ^2 x}{2}}\right) d x$
$=\frac{1}{2} \int x \operatorname{cosec} \frac{x}{2} d x+\int \frac{\frac{\cos x}{2}}{\frac{\sin x}{2}} d x$
$=\frac{1}{2}\left[x \int \operatorname{cosec} \frac{x}{2} d x-\int\left(\frac{ d }{ d x}(x) \int \operatorname{cosec} \frac{x}{2} d x\right) d x\right]+\int \cot \frac{x}{2} d x$
$=\frac{1}{2}\left[x\left(\frac{-\cot \frac{x}{2}}{\frac{1}{2}}\right)-\int 1 \cdot\left(\frac{-\cot \frac{x}{2}}{\frac{1}{2}}\right) d x\right]+\int \cot \frac{x}{2} d x$
$=\frac{1}{2}\left(-2 x \cot \frac{x}{2}+2 \int \cot \frac{x}{2} d x\right)+\int \cot \frac{x}{2} d x$
$=-x \cot \frac{x}{2}+\int \cot \frac{x}{2} d x+\int \cot \frac{x}{2} d x$
$=-x \cot \frac{x}{2}+4 \log \left|\sin \left(\frac{x}{2}\right)\right|+ c$
$=-x \cot \frac{x}{2}+2 \cdot \frac{\log \left|\sin \left(\frac{x}{2}\right)\right|}{\frac{1}{2}}+ c $
$=\int\left(\frac{x+2 \frac{\sin x}{2} \frac{\cos x}{2}}{2 \frac{\sin ^2 x}{2}}\right) d x$
$=\int\left(\frac{x}{2 \frac{\sin ^2 x}{2}}+\frac{2 \frac{\sin x}{2} \frac{\cos x}{2}}{2 \frac{\sin ^2 x}{2}}\right) d x$
$=\frac{1}{2} \int x \operatorname{cosec} \frac{x}{2} d x+\int \frac{\frac{\cos x}{2}}{\frac{\sin x}{2}} d x$
$=\frac{1}{2}\left[x \int \operatorname{cosec} \frac{x}{2} d x-\int\left(\frac{ d }{ d x}(x) \int \operatorname{cosec} \frac{x}{2} d x\right) d x\right]+\int \cot \frac{x}{2} d x$
$=\frac{1}{2}\left[x\left(\frac{-\cot \frac{x}{2}}{\frac{1}{2}}\right)-\int 1 \cdot\left(\frac{-\cot \frac{x}{2}}{\frac{1}{2}}\right) d x\right]+\int \cot \frac{x}{2} d x$
$=\frac{1}{2}\left(-2 x \cot \frac{x}{2}+2 \int \cot \frac{x}{2} d x\right)+\int \cot \frac{x}{2} d x$
$=-x \cot \frac{x}{2}+\int \cot \frac{x}{2} d x+\int \cot \frac{x}{2} d x$
$=-x \cot \frac{x}{2}+4 \log \left|\sin \left(\frac{x}{2}\right)\right|+ c$
$=-x \cot \frac{x}{2}+2 \cdot \frac{\log \left|\sin \left(\frac{x}{2}\right)\right|}{\frac{1}{2}}+ c $
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