Question

If $\int \frac{1}{1+\sin x}=-\frac{2}{(a+\tan \frac{x}{2})}+C$, then

• A. $a=1,C\in R$
• B. $a=\frac{\pi }{4},C\in R$
• C. $a=-1,C\in R$
• D. $a=-\frac{\pi }{3},C\in R$

Answer 1

Answer: (A)

$a=1,C\in R$

We have,

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

$I=\int \frac{dx}{1+2\sin \frac{x}{2}\cos \frac{x}{2}}$

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

$I=\int \frac{{\sec }^2\frac{x}{2}dx}{({\sec }^2\frac{x}{2}+2\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}})}$

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

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

Let $t=1+\tan \frac{x}{2}$

$\frac{dt}{dx}=0+{\sec }^2\frac{x}{2}\times \frac{1}{2}$

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

$I=2\int \frac{dt}{{\left(t\right)}^2}$

$I=2(-\frac{1}{t})+C$

$I=-\frac{2}{(1+\tan \frac{x}{2})}+C$

On comparing, we get

$a=1,C\in R$.

Hence, this is the answer.