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

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

Using $x=3sin\theta$

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

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

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

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

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

But don't know what to do next.

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

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

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

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

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

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

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

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

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

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

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

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

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 $x=3sin\theta$

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

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

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

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

where it's sufficient to take the positive root.