Question

Find the integrals of the function: $sin^3 (2x+1)$

Answer 1

The correct answer is: $=\frac{-cos(2x+1)}{2}+\frac{cos^3(2x+1)}{6}+C$
Let $I = ∫sin^3(2x+1)$
$⇒ ∫sin^3(2x+1)dx = ∫sin^2(2x+1).sin(2x+1)dx$
$= ∫(1-cos^2(2x+1)sin(2x+1)dx$
Let $cos(2x+1)=t$
$⇒ -2sin(2x+1)dx = dt$
$⇒ sin(2x+1)dx = \frac{-dt}{2}$
$⇒ I = \frac{-1}{2} ∫(1-t^2)dt$
$=-\frac{1}{2}[t-\frac{t^3}{3}]$
$=\frac{-1}{2}{cos(2x+1)-\frac{cos^3(2x+1)}{3}}$
$=\frac{-cos(2x+1)}{2}+\frac{cos^3(2x+1)}{6}+C$