# What did I do Wrong with this integration by u-sub

1. Jul 10, 2014

### shreddinglicks

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

∫x^2(a^2-x^2)^.5 limits from 0 to a
2. Relevant equations

∫x^2(a^2-x^2)^.5 limits from 0 to a

3. The attempt at a solution

It's way to much for me to type in the short amount of time I have so I included a picture of my work. It's neat and easy to read.

the answer should be (PI/16)*a^4

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2. Jul 10, 2014

### LCKurtz

To tell you the truth, I didn't read the whole thing. But I suggest where you have$$a^4\int_0^a \sin^2\theta \cos^2\theta~d\theta$$you write that as$$a^4\int_0^a \left(\frac {\sin(2\theta}{2}\right )^2~d\theta$$and immediately use the double angle formula for the $\sin^2(2\theta)$. It will be much easier.

3. Jul 10, 2014

### shreddinglicks

Which Identity did you use to get that?

4. Jul 10, 2014

### LCKurtz

$\sin(2\theta)=2\sin\theta\cos\theta$

5. Jul 10, 2014

### Pranav-Arora

You did not change the limits when you did the substitution.

$$a^4\int_0^{\pi/2}\sin^2\theta \cos^2\theta\,d\theta=\frac{a^4}{4}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta$$

6. Jul 10, 2014

### shreddinglicks

Why does my a become PI/2?

7. Jul 10, 2014

### Pranav-Arora

$x=\sin\theta$, at $x=a$, $a=a\sin\theta \Rightarrow \theta=\pi/2$.

8. Jul 10, 2014

### shreddinglicks

I understand that. The problem is when I apply it, it just gives me my original integral.

9. Jul 10, 2014

### shreddinglicks

I don't see how that matters. a/a = 1 and the arcsin(1) = PI/2

It's the same.

10. Jul 10, 2014

### shreddinglicks

I know it's a lot to read but can someone please look over my entire work? I am very close to the answer except for the numerator.

11. Jul 10, 2014

### LCKurtz

Apparently you don't understand it. Use$$\sin^2(2\theta) = \frac {1-\cos(4\theta)}2$$And, as another poster has noticed and I missed, the upper limit for $\theta$ should be $\frac \pi 2$.

12. Jul 11, 2014

### LCKurtz

So, are you going to try it? It's only one simple step left to integrate to the correct answer.

13. Jul 11, 2014

### shreddinglicks

I did get the answer. I went to sleep last night, where I'm located it was past midnight. I solved it this morning.

I didn't quite do it the way you guys have suggested.

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14. Jul 11, 2014

### LCKurtz

You should try it. Much simpler and you might learn something useful.