Integrals — Class 12 Maths Solution

: Let $I = \int {{{\sin }^2}\left( {2x + 5} \right)dx} = \cfrac{1}{2}\int {\left[ {1 - \cos 2\left( {2x + 5} \right)} \right]} dx$

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