Question

Evaluate integral:

∫cos³xsinx dx

Answer 1

\(\int cos^3x \,sinx\,dx = - \int t^3 dt\)   \(\begin{pmatrix}\because \text{Let} \, cos x = t\\⇒-sinx\,dx = dt\\⇒sinx\,dx = - dt\end{pmatrix}\) 

\(= \frac{-t^4}4 + C\)

\(= \frac{-cos^4x}{4}+C\)      \((\because t = cos x)\)