# Math Physics Tricky Integrals go Over my Head

1. Feb 20, 2008

### dazednconfuze

so i have this question about a ladder and the area under it and when it all comes down to it I get the integral, ∫ u2 ( 1 – u2 ) 3/2 du, from u = 0 to u = 1.

I am not sure whether to use trig substitution. I keep ending up in the same boat when I do. . . And the same thing when I do integration by parts. I'm not too quick with integration obviously. any help?

2. Feb 20, 2008

### blochwave

is u2 u*2?

I'm almost gonna have to guess it's u*(1-u)^(-3/2) that's giving you trouble(or 2*u or whatever)

in which case I think integration by parts will do it

What's the actual problem though?

3. Feb 20, 2008

### clem

I assume you mean $$\int_0^1 u^2(1-u^2)^{3/2} du$$.
Let $$u=\sin\theta$$ and brush up on your trig.
Use the formulae for sin and cos of theta/2.

4. Feb 20, 2008

### blochwave

oh ok, that makes more sense with what he typed >_>

You'll need those trig identities, but you can find huge tables of such identities in moments courtesy of the internet

5. Feb 20, 2008

### dazednconfuze

I tried substituting u = sin(theta) and all I got was sin^2(theta) cos^4(theta) dtheta... and that just doesn't mean anything to me...

6. Feb 20, 2008

### dextercioby

$$\int \sin^2 \theta {} \cos^4 \theta {} {} d\theta = \frac{1}{4}\int \sin^2 2\theta \ \frac{1+\cos 2\theta}{2} {} d\theta$$

Can you carry on from here ?