# Integration problems: Intriguing Integrals

1. Apr 2, 2008

### BioCore

1. The problem statement, all variables and given/known data
$$\int$$x^3$$\sqrt{}a^2-x^2$$dx

2. Relevant equations

3. The attempt at a solution
I came this far with my solution:

a^5$$\sqrt{}sin^3\Theta cos^2\Theta$$d$$\Theta$$
a^5$$\sqrt{}sin^3\Theta-sin^5\Theta$$d$$\Theta$$

The answer given to me is:

-1/15(a^2 - x^2)^3/2 (2a^2 + 3x^2) + C

It seems that they used sine theta, and not cos theta. I have currently tried to do that but am not sure how I would get rid of the extra sin$$\Theta$$.

Thanks for the help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Apr 2, 2008
2. Apr 2, 2008

### rock.freak667

$$\int x^3 \sqrt{a^2-x^2}dx$$

let $x=acos\theta \Rightarrow dx=-asin\theta$

$$\int x^3 \sqrt{a^2-x^2}dx \equiv \int (acos\theta)^3\sqrt{a^2-a^2cos^2\theta} \times -asin\theta d\theta$$

$$\int - a^4cos^3\theta \sqrt{a^2(1-cos^2\theta)} \times -asin\theta d\theta$$

$$-a^5 \int cos^3\theta sin^2\theta d\theta$$

recall that $sin^2\theta +cos^2\theta=1$

and substitute for $sin^2\theta$

3. Apr 2, 2008

### BioCore

I would assume that using your solution I would have the right answer. The only problem is that in the solutions my Professor gave us, he has an intermediate step before the final solution such as this:

$$a^5 \int sin^3\theta cos^2\theta d\theta$$

Also in explaining to us Trig Substitution he placed $$\sqrt{a^2 - x^2}$$ as the bottom portion of the right triangle. Thus he stated that

x = asin$$\Theta$$
dx = acos$$\Theta$$
$$\sqrt{a^2 - x^2}$$ = acos\theta

Last edited by a moderator: Apr 2, 2008
4. Apr 2, 2008

### rock.freak667

Your teacher used x=sin$\theta$, I used x=cos$\theta$.
So in his triangle, $sin\theta=\frac{x}{a}$ which would make the adjacent side,$\sqrt{a^2-x^2}$. At the end of that integral you would get terms involving $cos\theta$

In my substitution, the opposite side would be $\sqrt{a^2-x^2}$ and at the end of my integral I would have terms involving $sin\theta$

So in theory it should all work out the same.

5. Apr 2, 2008

### BioCore

$$-a^5 \int \sqrt{1 - sin^2\theta} (1 - sin^2\theta) (sin^2\theta)d\theta$$

= $$-a^5 \int (1 - sin^2\theta)^3/2 (sin^2\theta)d\theta$$

= $$-a^5 \int sin^3\theta - sin^6\theta d\theta$$

= $$-a^5 [1/4sin^4\theta + 1/7sin^7\theta ]$$ +C

Now I know that I should substitute for the sine function, but is this so far correct?