# How to do it faster?

1. Apr 2, 2005

### bomba923

(excuse my poor LaTex...i don't know it very well yet )
$$2\pi \int {x^3 \sin 2x\,dx} \Rightarrow \left\{ \begin{array}{l} u = 2x \\ du = 2dx \\ \end{array} \right\} \Rightarrow \frac{\pi }{8}\int {u^3 \sin u\,du} = \frac{\pi }{8}\left( { - u^3 \cos u + \int {u^2 \cos u\,du} } \right)$$
$$= \frac{\pi }{8}\left( { - u^3 \cos u + u^2 \sin u + u\cos u - \sin u} \right)$$
$$= \frac{{\pi \left[ { - 8x^3 \cos \left( {2x} \right) + 4x^2 \sin \left( {2x} \right) + 2x\cos {2x} - \sin {2x}} \right]}}{8}$$

How can I do this faster? Are there things I can skip or connect--etc--?

Last edited: Apr 2, 2005
2. Apr 3, 2005

### Muzza

The substitution u = 2x seems unnecessary.

3. Apr 3, 2005

### p53ud0 dr34m5

just use tabular integration. that would be the fastest way to do it.

4. Apr 3, 2005

### bomba923

Thanx--I tried it and it took much less time

5. Apr 3, 2005

### p53ud0 dr34m5

no problem. glad i could help.