Integrals — Maths STD 12 Science — Question

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

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

$= \int {\left( {1 - \frac{1}{{2{{\cos }^2}\frac{x}{2}}}} \right)} dx$

As $2{\cos ^2}\frac{\theta }{2} = 1 + \cos \theta$

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

$= x - \frac{1}{2}\frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}} + c$ $[\because\int {{{\sec }^2}\left( {ax + b} \right)} dx = \frac{{\tan \left( {ax + b} \right)}}{a} + c]$

$= x - \tan \frac{x}{2} + c$