# Integral Problem (very Hard)

1. Mar 4, 2010

### _wolfgang_

1. The problem statement, all variables and given/known data
Prove that $$\int$$$$\sqrt{}9-x^2$$ dx

=$$\frac{9\Theta}{2}$$+$$\frac{9sin2\Theta}{4}$$+c

given that x=3sin$$\Theta$$
2. Relevant equations

3. The attempt at a solution
$$\int$$$$\sqrt{}9-x^2$$ dx

=$$\frac{(9-x^2)^{1.5}}{10x}$$

=$$\frac{(-x^2)(\sqrt{-x^2+9}+9\sqrt{(-x^2)+9}}{10x}$$

=$$\frac{-x(x^2\sqrt{(-x^2)+9}-9(\sqrt{(-x^2)+9}}{10}$$

=$$\frac{-3sin\Theta(9(sin^2)\Theta\sqrt{9(sin^2)\Theta-9}-9\sqrt{9(sin^2)+9}}{10}$$

Im not to sure if im going in the right direction if i am not guidance would be appreciated

2. Mar 4, 2010

### vela

Staff Emeritus
You're not going in the right direction. Your very first step is wrong. The power rule only applies when the integrand is of the form xndx where $n \ne -1$. It doesn't apply when you have some function of x taken to a power, as you do in this case. Also, I have no idea where that 10x in the denominator came from.

Use the substitution given and rewrite the integral in terms of $\theta$ first.

3. Mar 4, 2010

### _wolfgang_

ok so if i do...

$$\int$$$$\sqrt{9-9sin^2\Theta}$$
=$$\int$$3-3sin$$\Theta$$
=3$$\Theta$$+3cos$$\Theta$$

is it going in the right direction now?? thanks for the help

Last edited: Mar 4, 2010
4. Mar 4, 2010

### vela

Staff Emeritus
Marginally better. First of all,

$$\sqrt{a^2-b^2} \ne a-b$$

Second, you forgot the dx and then didn't write it in terms of $d\theta$. Finally, not that it really matters, you didn't integrate the first term correctly.

I would suggest you review your textbook on the topic of trig substitutions. There's probably a similar example you could use as a template for solving this problem.

5. Mar 4, 2010

### HallsofIvy

$\sqrt{9- 9 sin^2(\theta)}= 3\sqrt{1- sin^2(\theta)}= 3\sqrt{cos^2(\theta)}$

6. Mar 4, 2010

### _wolfgang_

okay i was finally able to prove it!!

$$\int\sqrt{a^2-x^2}$$ dx
=$$\int a^2-a^2sin^2\Theta$$ acos$$\Theta$$ d$$\Theta$$
=$$\int\sqrt{a^2(1-sin^2\Theta}$$ acos$$\Theta$$ d$$\Theta$$
=$$\int\sqrt{a^2cos^2\Theta}$$ acos$$\Theta$$ d$$\Theta$$
=$$\int a^2cos^2\Theta$$ d$$\Theta$$

=$$\frac{a^2}{2}$$($$\Theta$$+sin$$\Theta$$cos$$\Theta$$)

i know i definatly skipped a couple of steps in integrating cos^2

Last edited: Mar 4, 2010