Find $\int\sin^3(2x+1)dx$.

Question

Find $\int\sin^3(2x+1)dx$. Having three different results which one is right right?

1.$y= -\frac{\cos (2x+1)}{2} +\frac{1}{24\cos(6x^3)} -\frac{1}{4\cos(2x+1)}+c$

2. $y=-\cos (2x+1) + \cos^3(2x+1) +c.$

3. $y= -\frac{\cos(6x+3)}{24} -\frac{3}{8}\cos(2x+1) +c$

Answer 1

Hints

1. Substitute $u = 2x + 1$. What will $du$ be?
2. Use the reduction formula.

Answer 2

Hint: $${\displaystyle \sin ^{3}x={\frac {3\sin x -\sin(3x )}{4}}}$$ and $${\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos(3\theta )}{4}}}$$