Question

The value of $\int \frac{2+\sin x}{1+\cos x}{e}^{x/2}dx$ is

• A. $2.e^{x/2}\tan \frac{x}{2}+C$
• B. $e^{x/2}\tan x+C$
• C. $\frac{1}{2}{e}^{x/2}\sin x+C$
• D. $\frac{1}{2}{e}^{x/2}\sin \frac{x}{2}+C$

Answer 1

Answer: (A)

Let $l=\int \frac{2+\sin x}{1+\cos x}.e^{x/2}dx$
$\Rightarrow$ $l=\int \frac{2+\frac{2\tan x/2}{1+{\tan }^{2}x/2}}{1+\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}}.e^{-x/2}dx$
$\Rightarrow$ $l=\frac{2{\tan }^2\frac{x}{2}+2+2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}-{\tan }^2\frac{x}{2}+1}.e^{x/2}dx$
$\Rightarrow$ $l=2\int \frac{{\tan }^2\frac{x}{2}+\tan \frac{x}{2}+1}{2}.e^{x/2}dx$
$\Rightarrow$ $l=\int {\tan }^2\frac{x}{2}.e^{x/2}dx+\int \tan \frac{x}{2}.e^{x/2}dx$
$+\int e^{x/2}dx$
$\Rightarrow$ $l=\int \underset{II}{{\sec }^2}x/2.e^{x/2}dx$
$-\int e^{x/2}dx+\int \tan \frac{x}{2}.e^{x/2}dx+\int e^{x/2}dx$
$\Rightarrow$ $l=2e^{x/2}.\tan \frac{x}{2}-\int \frac{1}{2}{e}^{x/2}.\tan \frac{x}{2}.2dx$
$+\int \tan \frac{x}{2}.e^{x/2}dx+C$
$\Rightarrow$ $l=2e^{x/2}.\tan \frac{x}{2}-\int e^{x/2}.\tan \frac{x}{2}dx$
$+\int e^{x/2}.\tan \frac{x}{2}dx+C$
$\Rightarrow$ $l=2e^{x/2}.\tan \frac{x}{2}+C$