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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 20 Maths Textbook Solution.

$-e^{x} \cot \frac{x}{2}+C$

Given:

$\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$

Hint:

You must know about the derivation of cot x and $\int e^{x} d x$

Explanation:

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

$=\int e^{x}\left(\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos _{2}^{x}}{2 \sin ^{2} \frac{x}{2}}\right) d x$                                                      $\left[\begin{array}{l} \because \sin ^{2} \theta+\cos ^{2} \theta=1 ; \sin 2 \theta=2 \sin \theta \cos \theta ; \\ 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2} \end{array}\right]$

\begin{aligned} &=\frac{1}{2} \int e^{x}\left(\frac{\sin {\frac{x}{2}}-\cos \frac{x}{2}}{\sin \frac{x}{2}}\right)^{2} d x \\ &=\frac{1}{2} \int e^{x}\left(1-\cot \frac{x}{2}\right)^{2} d x \\ &=\frac{1}{2} \int e^{x}\left(\left(1+\cot ^{2} \frac{x}{2}\right)-2 \cot \frac{x}{2}\right) d x \end{aligned}                                                                            $\left[\because \cot x=\frac{\cos x}{\sin x}\right]$

$=\frac{1}{2} \int e^{x}\left(\operatorname{cosec}^{2} \frac{x}{2}-2 \cot \frac{x}{2}\right) d x$                                                                                           $\left[\because 1+\cot ^{2} \theta=\cos e^{2} \theta\right]$

\begin{aligned} &=\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x-\int \cot \frac{x}{2} e^{x} d x \\ =&-\int \cot \frac{x}{2} \cdot e^{x} d x+\int \cos e c^{2} \frac{x}{2} e^{x} \cdot \frac{1}{2} d x \\ =&-\left[\cot \frac{x}{2} \cdot e^{x}-\int-\operatorname{cosec}^{2} \frac{x}{2} \cdot \frac{1}{2} \cdot e^{x} d x\right]+\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x \ldots . .\left[\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right] \\ =&-\cot \frac{x}{2} \cdot e^{x}-\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x+\frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x \\ =&-\cot \frac{x}{2} \cdot e^{x}+C \end{aligned}