# Homework Help: Trignometric Substiution Problem

1. Feb 9, 2013

### Archimedes II

1. The problem statement, all variables and given/known data

$\displaystyle \int _0 ^a 4x^2 \sqrt{a^2-x^2} dx$

2. Relevant equations

$\cos^2 x =1 - \sin^2 x$

3. The attempt at a solution

Substuting

Substituting $x=a \sin(\theta)$

$dx=a\cos( \theta) d\theta$

x=0
asin$\theta$=0
sin$\theta$=0
$\theta$=0

$\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =$

$\displaystyle \int_{0}^{\pi/2} 4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=$

$\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta \sqrt{a^2-a^2sin^2\theta}a\cos\theta d\theta=$

$\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a \sqrt{1-sin^2\theta}a\cos\theta d\theta=$

$\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a cos\theta a\cos\theta d\theta=$

$\displaystyle 4a^4\int_{0}^{\pi/2} sin \theta cos^2 \theta d\theta=$

U substitution

$u=cos\theta$

$-du= sin\theta d\theta$

$\displaystyle 4a^4\int u^2 du=$
When I put in the bounds it yields from 1 to 0. What did I do wrong

Last edited: Feb 9, 2013
2. Feb 9, 2013

### Archimedes II

So the code is messed up basically

Integral from 0 to a [4x^2 squareroot(a^2 - x^2)]dx =

4a^4 integral from 0 to pi/2 [sin theta cos^2 theta] dtheta

I'm stuck trying to change the bounds with a u substitution

Last edited: Feb 9, 2013
3. Feb 9, 2013

### SammyS

Staff Emeritus
I fixed your LaTeX code up to this point.

Here is the Code for some of the lines I fixed:

Code (Text):
$\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =$

$\displaystyle \int_{0}^{\pi/2} 4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=$

The "\displaystyle" modifier allows the integral symbol to be larger than otherwise.

I'll let you edit the Original Post .

Last edited: Feb 9, 2013
4. Feb 9, 2013

### Archimedes II

Ok, I fixed it thank you very much. I didn't know how to put a bound into the intergral code.

5. Feb 9, 2013

### SammyS

Staff Emeritus
So you get

$\displaystyle -4a^4\int_{1}^{0} u^2\,du$

Right?

6. Feb 9, 2013

### Archimedes II

Yes but the how can the bound be from a higher number to a smaller one.

7. Feb 9, 2013

### SammyS

Staff Emeritus
What does switching the order do to the integral?

8. Feb 9, 2013

### Archimedes II

I solved it I forgot to square the sin at the end. I did't need to use a u substitution.

Any regardless if you have that bound it yields a negative answer.

Thanks for helping though.

$\pi a^4/4$

9. Feb 9, 2013

### SammyS

Staff Emeritus
Yes. It works out just fine either way.

Switching the limits of integration switches the sign. That's good because it got rid of that leading negative sign.

... and you learned a bit more of LaTeX coding .