Question

Evaluate $I=\int \frac{e^x(1+\sin x)+e^{-x}(1-\sin x)}{1+\cos x}dx$

• A. $(e^x-e^{-x})\tan \frac{x}{2}+c$
• B. $(e^x-e^{-x})\text{cosec}\frac{x}{2}+c$
• C. $(e^x-e^{-x})\cot \frac{x}{2}+c$
• D. $(e^x-e^{-x})\sec \frac{x}{2}+c$

Answer 1

The correct option is D $(e^x-e^{-x})\sec \frac{x}{2}+c$

$I=\int \frac{e^x(1+\sin x)+e^{-x}(1-\sin x)}{1+\cos x}dx$
We know that
$\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$
$\Rightarrow I=\int \frac{e^x-e^{-x}+(e^x-e^{-x})\sin x}{1+\cos x}dx\Rightarrow I=\int [\frac{e^x+e^{-x}}{2}{\sec }^2\frac{x}{2}+(e^x-e^{-x}\left)\tan \frac{x}{2}\right]dx$
$\Rightarrow I=\frac{1}{2}\left[\int e^x({\sec }^2\frac{x}{2}+2\tan \frac{x}{2})dx+\int e^{-x}({\sec }^2\frac{x}{2}-2\tan \frac{x}{2})dx\right]$
$\Rightarrow I=(e^x-e^{-x})\tan \frac{x}{2}+c$