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.