Question

Evaluate the definite integrals.

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

Answer 1

$LetI={\int }_{\frac{\pi }{2}}^{\pi }{e}^x\left(\frac{1-sinx}{1-cosx}\right)dx={\int }_{\frac{\pi }{2}}^{\pi }{e}^x\left(\frac{1-2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}{2si{n}^2\left(\frac{x}{2}\right)}\right)dx(\because sinx=2sin\frac{x}{2}cos\frac{x}{2}and1-cosx=2si{n}^2\frac{x}{2})={\int }_{\frac{\pi }{2}}^{\pi }{e}^x(\frac{1}{2}cose{c}^2\frac{x}{2}-cot\frac{x}{2})dx={\int }_{\frac{\pi }{2}}^{\pi }{e}^x(-cot\frac{x}{2}+\frac{1}{2}cose{c}^2\frac{x}{2})dx\left[\begin{array}{c}Here,\frac{d}{dx}(-cot\frac{x}{2})=\frac{1}{2}cose{c}^2\frac{x}{2}itistheformof\\ \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)\end{array}\right]\therefore I={\left[e^x(-cot\frac{x}{2})\right]}_{\frac{\pi }{2}}^{\pi }=-e^{\pi }cot\left(\frac{\pi }{2}\right)-[-e^{\frac{\pi }{2}}cot\left(\frac{\pi }{4}\right)]=-e^{\pi }.0+e^{\frac{\pi }{2}}.1=e^{\frac{\pi }{2}}$