# Ideas as to how I should start to integrate this one?

1. Sep 24, 2005

### LocationX

$$\int (C^2x^2-x^4)^{0.5} dx$$

C is a constant

any ideas as to how I shoud start to integrate this one?

2. Sep 24, 2005

### LeonhardEuler

Yeah, just factor out x^2, so you have
$$\int x\sqrt{C^2-x^2}dx$$
Then just make the substitution $u=x^2$.

3. Sep 24, 2005

### LocationX

so then how would I integrate

$$\int (C^2-u)^{0.5}$$

4. Sep 25, 2005

### Luminous Blob

You can make another substitution, e.g. $$z = C^2 - u$$ or just make your original substitution $$u = C^2 - x^2$$, which gives

$$du = -2x dx$$

$$\frac{du}{-2x} = dx$$

then you just use your normal integration rules.

5. Sep 25, 2005

### arildno

Just a minor correction.
We should have:
$$\int\sqrt{C^{2}x^{2}-x^{4}}dx=\int|x|\sqrt{C^{2}-x^{2}}dx$$