# Integration help

1. Apr 12, 2005

### mkkrnfoo85

I was doing a change in variables problem in multivariable calculus and I got stuck on the last integration.

$$\frac {4}{3} \int_{u=0}^{1} (1-u^2)^{\frac {1}{2}}(2u^2+1)du$$

I don't think substitution works. Can anyone show me an easy way to solve this? Thanks.

2. Apr 12, 2005

### dextercioby

Typically $u=\sin t$

Daniel.

3. Apr 13, 2005

### Theelectricchild

For problems like this in general, you should try spliting up any integrand into as many pieces as possible...

Here you can rewrite the integrand obviously as 2u^2[(1-u^2)^0.5] + (1-u^2)...

and being the lazy engineering student that I am... I would look these up in a table of integrals--- which would definetely have the general solutions.

4. Apr 13, 2005

### dextercioby

Don't listen to an engineering student giving advice in anything but engineering...

The transformed integral should be

$$\frac{4}{3}\int_{0}^{\frac{\pi}{2}} \cos^{2}t\left(2\sin^{2}t+1\right) dt$$

Use the double angle formulas to get it simplified.

Daniel.

5. Apr 13, 2005

### Phymath

did u miss $$(1-u)^{1/2}$$ -> $$\int cos t (2 sin^2 t + 1) dt$$? Or did i miss something

6. Apr 13, 2005

### dextercioby

Yeah,u missed the second cosine.One from the sqrt & the other from the change of variable...

Daniel.

7. Apr 13, 2005

### mkkrnfoo85

ah that clears things up. thanks.