Evaluate : \(\displaystyle\int x^2\sin3x\,dx\)

Answer: \(-\dfrac{x^2\cos3x}{3}+\dfrac{2x\sin3x}{9}+\dfrac{2\cos3x}{27}+c\)

Integrating by parts (\(u=x^2\), \(v=\sin3x\)):

\(I=x^2\displaystyle\int\sin3x\,dx-\int\left[\dfrac{d}{dx}(x^2)\int\sin3x\,dx\right]dx=-\dfrac{x^2\cos3x}{3}+\dfrac23\int x\cos3x\,dx\).

Again by parts: \(\displaystyle\int x\cos3x\,dx=\dfrac{x\sin3x}{3}-\int\dfrac{\sin3x}{3}dx=\dfrac{x\sin3x}{3}+\dfrac{\cos3x}{9}\).

\(\therefore I=-\dfrac{x^2\cos3x}{3}+\dfrac23\left[\dfrac{x\sin3x}{3}+\dfrac{\cos3x}{9}\right]+c=-\dfrac{x^2\cos3x}{3}+\dfrac{2x\sin3x}{9}+\dfrac{2\cos3x}{27}+c\).