Question

${\int }_{\frac{\pi }{2}}^{\pi }{e}^x\left(\frac{1-sinx}{1-cosx}\right)dx$

Answer 1

${\int }_{\pi /2}^{\pi }{e}^x\left(\frac{1-\sin x}{1-\cos x}\right)dx$

Let $I=\int e^x\left(\frac{1-\sin x}{1-\cos x}\right)dx$

or $I=\int e^x\frac{1-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx$

or, $I=\int e^x[\frac{{\csc }^2\frac{x}{2}}{2}-\cot \frac{x}{2}]dx$

Let $f\left(x\right)=-\cot \left(\frac{x}{2}\right)$

$\Rightarrow f^{\prime }\left(x\right)=\frac{{\csc }^2\frac{x}{2}}{2}$

or, $I=\int e^x[f^{\prime }\left(x\right)+f\left(x\right)]dx$

or, $I=e^x\int f^{\prime }\left(x\right)dx-\int \frac{d}{dx}\left(e^x\right)\left(f^{\prime }\left(x\right)dx\right)dx+\int e^{x}f\left(x\right)dx$

$\Rightarrow I=e^{x}f\left(x\right)-\int e^{x}f\left(x\right)dx+\int e^{x}f\left(x\right)dx$

$\Rightarrow I=e^{x}f\left(x\right)+c$

$\Rightarrow I=-e^x\cot \left(\frac{x}{2}\right)+c$

$={\int }_{\pi /2}^{\pi }{e}^x\left(\frac{1-\sin x}{1-\cos x}\right)dx$

$={\left[{-e}^x\cot \frac{x}{2}\right]}_{\pi /2}$

$={-e}^{\pi }.0+{-e}^{\pi /2}\cot \frac{\pi }{4}$

$=e^{\pi /2}$

Hence, the answer is $e^{\pi /2}.$