`intsin^3(2x+1)dx`

Explanation

I=`intsin^3(2x+1)dx`

We know that `sin3θ=1/4​[3sinθ−sin^3θ]`

I=`intsin^3(2x+1)dx`

I=`int[3sin(2x+1)−sin3(2x+1)]dx`

Let t=2x+1⇒dt=2dx

I=`int[3sin(t)−sin3(t)]dx`

=`−3/8​cost+1/24​cos3t+c`

=`−3/8​cos(2x+1)+1/24​cos3(2x+1)+c`