Question

Find the integral of the function
${\sin }^2\left(2x+5\right)$

Answer 1

Consider the given integral.

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

Let $t=2x+5$

$\frac{dt}{dx}=2+0$

$\frac{dt}{2}=dx$

Therefore,

$I=\frac{1}{2}\int {\sin }^2\left(t\right)dt$

$I=\frac{1}{4}\int (1-\cos 2t)dt$

$I=\frac{1}{4}[t-\frac{\sin 2t}{2}]+C$

On putting the value of $t$, we get

$I=\frac{1}{4}[2x+5-\frac{\sin 2(2x+5)}{2}]+C$

$I=\frac{1}{4}[2x+5-\frac{\sin (4x+10)}{2}]+C$

Hence, this is the answer.