Question

If $\int \frac{1}{1+sinx}dx=tan(\frac{x}{2}+a)+c,$ then

• A. $a=-\frac{\pi }{4},CϵR$
• B. $a=\frac{\pi }{4},CϵR$
• C. $a=\frac{5\pi }{4},CϵR$
• D. $a=\frac{\pi }{3},CϵR$

Answer 1

The correct option is A $a=-\frac{\pi }{4},CϵR$

We have,

$\int \frac{1}{1+\sin x}dx$

$=\int \frac{1}{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}dx$

$=\int \frac{1+{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}}dx$

$=\int \frac{{\sec }^2\frac{x}{2}}{{(1+\tan \frac{x}{2})}^2}dx$

Let

$\tan \frac{x}{2}=t$

$\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$

$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$

$\int \frac{2dt}{{(1+t)}^2}$

$\Rightarrow \int 2{(1+t)}^{-2}dt$

$\Rightarrow 2\int {(1+t)}^{-2}dt$

$\Rightarrow 2\frac{{(1+t)}^{-2+1}}{-2+1}$

$\Rightarrow 2\frac{{(1+t)}^{-1}}{-1}$

$\Rightarrow \frac{-2}{(1+t)}$

Put $t=\tan \frac{x}{2}$

Then,

$\Rightarrow \frac{-2}{(1+\tan \frac{x}{2})}=\tan \frac{x}{2}+a$

Hence, this is the answer.

Hence, this is the answer.