Integration Help: Solving Last Problem in Multivariable Calculus

[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.

 

[dextercioby]

Typically u=\sin t

Daniel.

 

[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 definitely have the general solutions.

 

[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.

 

[Phymath]

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

 

[dextercioby]

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

Daniel.

 

[mkkrnfoo85]

ah that clears things up. thanks.