Question

$\int\left(\frac{2- \sin 2x}{1- \cos 2x}\right)e^{x}dx$ is equal to

• A. $- e^{x} \cot\, x+c$
• B. $ e^{x} \cot\, x+c$
• C. $ 2e^{x} \cot\, x+c$
• D. $ -2e^{x} \cot\, x+c$

Answer 1

Answer: (A)

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