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)

$2.{{e}^{x/2}}\,\tan \frac{x}{2}+C$

Answer 2

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