Can I integrate this using only one integration by parts?

[bomba923]

$$2\pi \int {x^3 \sin 2x\,dx} \Rightarrow \left\{ \begin{array}{l}<br /> u = 2x \\ <br /> du = 2dx \\ <br /> \end{array} \right\} \Rightarrow \frac{\pi }{8}\int {u^3 \sin u\,du} = \frac{\pi }{8}\left( { - u^3 \cos u + \int {u^2 \cos u\,du} } \right) = \frac{\pi }{8}\left( { - u^3 \cos u + u^2 \sin u - \int {u\sin u\,du} } \right) = \frac{\pi }{8}\left( { - u^3 \cos u + u^2 \sin u + u\cos u - \sin u} \right) = \frac{{\pi \left[ { - 8x^3 \cos \left( {2x} \right) + 4x^2 \sin \left( {2x} \right) + 2x\cos x - \sin x} \right]}}{8} [\tex]<br /> <br /> How can I do this faster? Are there things I can skip or connect--etc--?$$

 

[dextercioby]

3 times part integration would do it...

$$\int \left( x^3\sin 2x\right) dx=\allowbreak -\frac 12x^3\cos 2x+\frac 34x^2\sin 2x-\frac 38\sin 2x+\frac 34x\cos 2x +C$$

Daniel.