Evaluate integral:∫cos3xsinx dx

1.	Evaluate integral:∫cos3xsinx dx
Answer» \(\int cos^3x \,sinx\,dx = - \int t^3 dt\) \(\begin{pmatrix}\because \text{Let} \, cos x = t\\⇒-sinx\,dx = dt\\⇒sinx\,dx = - dt\end{pmatrix}\) \(= \frac{-t^4}4 + C\) \(= \frac{-cos^4x}{4}+C\) \((\because t = cos x)\)

Answer»

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

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

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

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