Question

Find the integral of the function
${\sin }^3\left(2x+1\right)$

Answer 1

Consider the given integral.

$I=\int {\sin }^3(2x+1)dx$

Let $t=2x+1$

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

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

Therefore,

$I=\frac{1}{2}\int {\sin }^3\left(t\right)dt$

$I=\frac{1}{2}\int \frac{(3\sin t-\sin 3t)}{4}dt$

$I=\frac{1}{8}[-3\cos t+\frac{\cos 3t}{3}]+C$

On putting the value of $t$, we get

$I=\frac{1}{8}[-3\cos (2x+1)+\frac{\cos 3(2x+1)}{3}]+C$

$I=\frac{1}{8}[\frac{\cos (6x+3)}{3}-3\cos (2x+1)]+C$

Hence, this is the answer.