Mark the correct alternative in each of the following:
Evaluate $\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$
A. $-e^{x} \tan \frac{x}{2}+C$
B. $-e^{x} \cot \frac{x}{2}+C$
C. $-\frac{1}{2} \mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{C}$
D. $-\frac{1}{2} e^{x} \cot \frac{x}{2}+C$
Given, $\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$
$=-\int e^{x}\left(\frac{\sin x}{1-\cos x}-\frac{1}{1-\cos x}\right) d x\left\{\int e^{x}\left[f(x)+f^{\prime}(x)\right]=e^{x} f(x)\right\}$
$\Rightarrow f(x)=\frac{\sin x}{1-\cos x} ; f^{\prime}(x)=-\frac{1}{1-\cos x}$
$=-e^{x}\left(\frac{\sin x}{1-\cos x}\right)$
$\because\left[\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}\right]$
$=-e^{x} \cot \frac{x}{2}+c$
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