Integrals — Maths STD 12 Science — Question

$= \int {\frac{1}{2}\left\{ {1 - \cos 2\left( {2x + 5} \right)} \right\}dx} $

Using ${\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}$

$ = \frac{1}{2}\int {\left\{ {1 - \cos \left( {4x + 10} \right)} \right\}dx} $

$= \frac{1}{2}\left[ {\int {1dx - \int {\cos \left( {4x + 10} \right)dx} } } \right]$

Using $\int {\cos \left( {ax + b} \right)dx = \frac{{\sin \left( {ax + b} \right)}}{a} + c} $

$= \frac{1}{2}\left[ {x - \frac{{\sin \left( {4x + 10} \right)}}{4}} \right] + c$

$= \frac{1}{2}x - \frac{1}{8}\sin \left( {4x + 10} \right) + c$