# Thread: help with a trig substution integral

1. ## help with a trig substitution integral

I'd appreciate if someone can help me out with the following integral:

$\displaystyle \int x*\sqrt{81-9x^2} dx$

I know this can be done with the power rule, but I'm looking to do it using trig substitution. I have a triangle set up, where sin theta = 3x/9, but get stuck halfway through the problem. Can anyone help me figure this one out? Thanks!

2. $\displaystyle \int \frac{x~dx}{\sqrt{81-9x^2}}$

Substitute $\displaystyle x = 3 \sin{\theta}$

$\displaystyle dx = 3 \cos{\theta} d \theta$

$\displaystyle \int \frac{ (3 \sin{\theta} ) ~ 3 \cos{\theta} ~d \theta }{ 9 \sqrt{1 - \sin^2 \theta} }$

$\displaystyle = \int \frac{ \sin{\theta} ~ \cos{\theta} ~d \theta }{ \cos{\theta} }$

$\displaystyle = \int \sin{\theta} ~ d{\theta}$
$\displaystyle = - \cos{\theta} + C$

$\displaystyle = - \sqrt{ 1 - (\frac{x}{3})^2} + C$

$\displaystyle = - \frac{ \sqrt{ 9 -x^2}}{3} + C$

3. I'd be interested in seeing how it can be done with the "power rule"!

If you are wondering how simplependulum saw so quickly that he needed x= 3 sin($\displaystyle \theta$), try factoring that "81" in the square root out:
$\displaystyle \int \frac{x dx}{\sqrt{81- 9x^2}}= \int\frac{xdx}{\sqrt{81(1- \frac{x^2}{9}})}$$\displaystyle =\int\frac{xdx}{9\sqrt{1- (\frac{x}{3})^2}}$ and to use "$\displaystyle cos(\theta)= \sqrt{1- sin^2(\theta)}$", we must have $\displaystyle sin(\theta)= \frac{x}{3}$ or $\displaystyle x= 3 sin(\theta)$.

4. Thanks for the help from above posters, I appreciate the explanations. However, the original problem I had trouble with was
$\displaystyle \int x\sqrt{81-9x^2} dx$,
with x times the sqrt, not x divided by the sqrt, which is part of what was tricky for me. (I think it was the problem I had with the Latex syntax when I originally posted that might have caused the mix-up.) So I've drawn my triangle, and I'm pretty sure that $\displaystyle x=3 sin(\theta)$ and $\displaystyle dx=3 cos(\theta) d\theta$. I think that in theory, I should be able to figure out the rest and replace everything back into the equation, but I'm just stuck and unable to work through to get an answer. Is there something I am doing wrong? If not, how should I work this to get an answer?

5. Originally Posted by psu1024
Thanks for the help from above posters, I appreciate the explanations. However, the original problem I had trouble with was
$\displaystyle \int x\sqrt{81-9x^2} dx$,
with x times the sqrt, not x divided by the sqrt, which is part of what was tricky for me. (I think it was the problem I had with the Latex syntax when I originally posted that might have caused the mix-up.) So I've drawn my triangle, and I'm pretty sure that $\displaystyle x=3 sin(\theta)$ and $\displaystyle dx=3 cos(\theta) d\theta$. I think that in theory, I should be able to figure out the rest and replace everything back into the equation, but I'm just stuck and unable to work through to get an answer. Is there something I am doing wrong? If not, how should I work this to get an answer?
Take out the factor of 9 from the square root. Now substitute $\displaystyle u = 9 - x^2$. (This works for both your question and for the incorrect interpretation of your question).