Question

Evalaute

${\int }^{\frac{}{}}\frac{1}{1+cosx}dx$

Answer 1

We have,

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

Now,

$\int \frac{1}{1+\cos x}$

$\Rightarrow \int \frac{1}{1+\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}$

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

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

$\Rightarrow \int \frac{{\sec }^2\frac{x}{2}}{2}dx$

$\Rightarrow \frac{1}{2}\int {\sec }^2\frac{x}{2}dx$

On integrating and we get,

$\Rightarrow \frac{1}{2}\frac{\tan \frac{x}{2}}{\frac{1}{2}}+C$

$\Rightarrow \tan \frac{x}{2}+C$

Hence, this is the answer.