Q76 / 120 · VITEEE 2009 · Mathematics
$\int\left(\frac{2-\sin 2 x}{1-\cos 2 x}\right) e^{x} d x$ is equal to
- Correct answerA$-\mathrm{e}^{x} \cot x+\mathrm{c}$
- B$\mathrm{e}^{x} \cot x+\mathrm{c}$
- C$2 \mathrm{e}^{x} \cot x+\mathrm{c}$
- D$-2 \mathrm{e}^{x} \cot x+\mathrm{c}$
Solution
$$\begin{aligned}
let I &=\int\left(\frac{2-\sin 2 x}{1-\cos 2 x}\right) e^{x} d x \\
&=\int\left(\frac{2-2 \sin x \cos x}{2 \sin ^{2} x}\right) e^{x} d x \\
&=\int \operatorname{cosec}^{2} x e^{x} d x-\int \cot x e^{x} d x \\
&=-\cot x e^{x}-\int(-\cot x) e^{x} d x \\
&-\int \cot x e^{x} d x+c \\
&=-\cot x e^{x}+c
\end{aligned}
$$
Hence, the correct answer is option: 'A'.