# Thread: [SOLVED] Trigonometric integration by parts #2

1. ## [SOLVED] Trigonometric integration by parts #2

Using $\displaystyle x=3sin\theta$

$\displaystyle \int x^3 \sqrt {9-x^2} ~dx$

I think I solved the part under the radical correctly ($\displaystyle 3cos\theta$)

If I plug in $\displaystyle x=3sin\theta$ and $\displaystyle dx=3cos\theta$ and use $\displaystyle 3cos\theta$ for the radical I get:

$\displaystyle \int 243sin^3\theta cos^2\theta ~d\theta$

I can change it to $\displaystyle 243\int sin\theta(1-cos\theta)(cos^2\theta)$

But don't know what to do next.

2. Originally Posted by redman223
Using $\displaystyle x=3sin\theta$

$\displaystyle \int x^3 \sqrt {9-x^2} ~dx$

I think I solved the part under the radical correctly ($\displaystyle 3cos\theta$)

If I plug in $\displaystyle x=3sin\theta$ and $\displaystyle dx=3cos\theta$ and use $\displaystyle 3cos\theta$ for the radical I get:

$\displaystyle \int 243sin^3\theta cos^2\theta ~d\theta$

I can change it to $\displaystyle 243\int sin\theta(1-cos\theta)(cos^2\theta)$

But don't know what to do next.
Make the substitution $\displaystyle u = \cos \theta$.

But you'd be much better off making the substitution $\displaystyle u = 9 - x^2$ in the original integral.

You get $\displaystyle - \frac{1}{2} \int (9 - u) \, \sqrt{u} \, du \, ......$

3. This is one of those problems where they want you to solve it using their method, so I am stuck using $\displaystyle x=3sin\theta$

Which when I set $\displaystyle u=cos\theta$ I come up with:

$\displaystyle -243(\frac {cos^3\theta}{3} - \frac {cos^4\theta}{4})+c$

If I am right, then what is the next step?

4. Originally Posted by redman223
This is one of those problems where they want you to solve it using their method, so I am stuck using $\displaystyle x=3sin\theta$

Which when I set $\displaystyle u=cos\theta$ I come up with:

$\displaystyle -243(\frac {cos^3\theta}{3} - \frac {cos^4\theta}{4})+c$

If I am right, then what is the next step?
Now you have to re-substitute $\displaystyle x = 3 \sin \theta$:

From the Pythagorean Identity, $\displaystyle \sin \theta = \frac{x}{3} \Rightarrow \cos \theta = \frac{\sqrt{9 - x^2}}{3} \, .....$

where it's sufficient to take the positive root.