#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 50 Maths Textbook Solution.

Answer:$e\frac{\pi }{2}$

Hint: Use indefinite integral formula then put the limits to solve this integral

Given:

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

Sol:

$\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x=\int_{\frac{\pi}{2}}^{\pi} e^{x} \frac{\left(1-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}\right)}{2 \sin ^{2} \frac{x}{2}} d x$

$\left[\begin{array}{l}\sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta\end{array}\right]$

$=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1}{2 \sin ^{2} \frac{x}{2}}-\frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}\right) d x$

$\left[\frac{1}{\sin \theta}=\operatorname{cosec} \theta, \cot \theta=\frac{\sin \theta}{\cos \theta}\right]$

$=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{\cos e c^{2} \frac{x}{2}}{2}-\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) d x$

$=\int_{\frac{\pi}{2}}^{\pi} \frac{e^{x} \cos e c^{2} \frac{x}{2}}{2}-\int_{\frac{\pi}{2}}^{\pi} e^{x} \cot \frac{x}{2} d x$

$=\frac{1}{2}\left[\int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2 \int_{\frac{\pi}{2}}^{\pi} e^{x} \cot \frac{x}{2} d x\right]$

$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{x}{2} \int e^{x} d x\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\left(\frac{d\left(\cot \frac{x}{2}\right)}{d x} \int e^{x} d x\right) d x\right\}$

Using integration by parts in second term,

$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{x}{2} e^{x}\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\left(-\cos e c^{2} \frac{x}{2}\right) \frac{1}{2} e^{x} d x\right\}$

$\left[\frac{d}{d x} \cot a x=-\cos e c^{2} a x \cdot a, \int e^{x} d x=e^{x}\right]$

$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{\pi}{2} \cdot e^{\pi}-\cot \frac{\pi}{4} e^{\frac{\pi}{2}}\right]+\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos e c^{2} \frac{x}{2} e^{x} d x\right\}$

$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-1\left(0 . e^{\pi}-1 \cdot e^{\frac{\pi}{2}}\right)-\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos e c^{2} \frac{x}{2} e^{x} d x$

$\left[\cot \frac{\pi}{2}=0, \cot \frac{\pi}{4}=1\right]$

$=-\left(0-e^{\frac{\pi}{2}}\right)$

$=e^{\frac{\pi}{2}}$