Question

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

• A. $e^{x} \cot x / 2$
• B. $e^{x} \tan x / 2$
• C. $-e^{x} \cot x / 2$
• D. $-e^{x} \tan x / 2$

Answer 1

Answer: (C)

$I=\int e^{x}\left(\frac{1-2 \sin x / 2 \cos x / 2}{2 \sin ^{2} x / 2}\right) d x$
$=\int e^{x}\left(\frac{1}{2} \text{cosec} x / 2-\cot x / 2\right) d x$
$=-\int e^{x}\left(\frac{1}{2} \cot x / 2-\frac{1}{2} \text{cosec}^{2}x / 2\right) d x$
$=-e^{x} \cot x / 2+c$