Question
Evaluate the definite integral $\int_{-\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$

Answer

Let $I=\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x\right.$ 
= $\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2}\left(\frac{x}{2}\right)}\right) d x\right.$ 
= $\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1}{2 \sin ^{2}\left(\frac{x}{2}\right)}-\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2}\left(\frac{x}{2}\right)}\right) d x\right.$ 
= $\int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)-\cot \frac{x}{2}\right) d x\right.$ 
Now let $f(x)=-\cot \frac{x}{2}$ 
$\Rightarrow$ f'(x) = $-\left(-\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)\right)=\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)$ 
$\Rightarrow \int_{-\frac{\pi}{2}}^{\pi}\left(e^{x}\left(\frac{1}{2} \csc ^{2}\left(\frac{x}{2}\right)-\cot \frac{x}{2}\right) d x\right.$ = $\int_{-\frac{\pi}{2}}^{\pi}\left(f(x)+f^{\prime}(x)\right) e^{x} d x$ 
= $\left[e^{x} f(x)\right]_{-\frac{\pi}{2}}^{\pi}$ 
= $\left[e^{x}\left(-\cot \frac{x}{2}\right)\right]_{-\frac{\pi}{2}}^{\pi}$ 
= $-\left[e^{\pi}\left(\cot \frac{\pi}{2}\right)-e^{\frac{\pi}{2}}\left(\cot \frac{\pi}{4}\right)\right]$
= $-\left[e^{\pi}(0)-e^{\frac{\pi}{2}}(1)\right]$ 
= $-\left[0-e^{\frac{\pi}{2}}\right]$ 
$\Rightarrow I=e^{\frac{\pi}{2}}$

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