Question

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

Answer 1

$I={\int }_{\frac{\pi }{2}}^{\pi }{e}^x\left(\frac{1-\sin x}{1-\cos x}\right)dx$
$={\int }_{\frac{\pi }{2}}^{\pi }{e}^x\left(\frac{1-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{2}}\right)dx$
$={\int }_{\frac{\pi }{2}}^{\pi }{e}^x(\frac{{\text{cosec}}^2\frac{\pi }{2}}{2}-\cot \frac{x}{2})dx$
Let $f\left(x\right)=-\cot \frac{x}{2}$
$\Rightarrow f^{\prime }\left(x\right)=-(-\frac{1}{2}{\text{cosec}}^2\frac{x}{2})=\frac{1}{2}{\text{cosec}}^2\frac{x}{2}$
$\therefore I={\int }_{\frac{\pi }{2}}^{\pi }{e}^x\left(f\right(x)+f^{\prime }(x\left)\right]dx$
$=[e^x\cdot f(x)dx{]}_{\frac{\pi }{2}}^{\pi }$
$=-{[e^x\cdot \cot \frac{x}{2}]}_{\frac{\pi }{2}}^{\pi }$
$=-[e^{\pi }\times \cot \frac{\pi }{2}-e^{\frac{\pi }{2}}\times \cot \frac{\pi }{4}]$
$=-[e^{\pi }\times 0-e^{\frac{\pi }{2}}\times 1]=e^{\frac{\pi }{2}}$