Question

$\int {\sin }^2(2x+5)dx$

Answer 1

$\int {\sin }^2(2x+5)dx$

$=\frac{1}{2}\int 2{\sin }^2(2x+5)dx$

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

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

Let $t=4x-10\Rightarrow dt=4dx$

$=\frac{1}{2}\int dx-\frac{1}{8}\int \cos tdt$

$=\frac{1}{2}x-\frac{1}{8}\sin t+c$

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

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