Evaluate the integrals \int sin^3(2x+1)dx

We know that sin3x = -4sin³x + 3sinx

-4sin³x = 3sinx - sin3x

\( sin^3x=\dfrac{3sinx - sin3x}{4}\)

\( \int \dfrac{3sin(2x+1) - sin3(2x+1)}{4}\)

\( \int sin\;ax\;dx=\dfrac{-1}{a}cos\;ax+c\)

On integrating we get,

\( \dfrac{-3}{8}cos(2x+1)+\dfrac{1}{24}cos(6x+3)+c\)