Question

Integrate: ${\sin }^3(2x+1)$

Answer 1

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

Let, $2x+1=t$

$2dx=dt$

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

We know that,

$\sin \left(3t\right)=3\sin t-4{\sin }^{3}t$

$4{\sin }^{3}t=3\sin t-\sin \left(3t\right)$

${\sin }^{3}t=\frac{1}{4}[3\sin t-\sin (3t\left)\right]$

$I=\frac{1}{2}\left[\int \frac{1}{4}\left(3\sin tdt\right)-\int \frac{1}{4}\sin \left(3t\right)dt\right]$

$I=\frac{1}{2}\times \frac{1}{4}[3\cos t-\frac{\cos 3t}{3}]+c$. substitute $t=2x+1$

$I=\frac{1}{24}\left[9\cos \right(2x+1)-\cos (6x+3\left)\right]+c$.