Integration Help: Solving x^2/sqrt(16-x^2) Integral

[VantagePoint72]

Hello,
I'm currently learning integration by trig substitution. The problem is:
integral ( x^2/sqrt(16-x^2) ) dx.
I seem to be having some difficulty...this is what I have:
let x=4sinu
dx = 4cosu du

Thus, we have:

integral 16sin^2(u)*4cosu du/(4cos^2(u)) (the last part is from the trig identity sin^2x + cos^2x = 1)

But now what? I simplied that as much as I could and then used integration by parts, but I ended up with the wrong answer. The correct answer is 8arcsin(x/4) - 0.5x*sqrt(16-x^2). Have I made a mistake in what I've done so far? If not, where do I proceed?

Thanks,
LOS

 

[Big-T]

$$\sqrt{1-\sin^2u}=|\cos u|$$.

 

[Gib Z]

Check your denominator again. Should the cosine term really be squared? Shouldn't be any need for integration by parts after that.

 

[flebbyman]

$$\int \frac{x^2}{\sqrt{16-x^2}}dx$$

$$x=4sin u, dx=4cos u du$$

$$\int\frac{16sin^2u}{(4cosu)}4cosu du$$

$$16\int sin^2udu$$

$$16(\frac{u}{2}-\frac{sinu*cosu}{2})$$

$$8u-8sinu*cosu$$

$$8sin^{-1}(\frac{x}{4})-2x\sqrt{1-(\frac{x}{4})^2}$$

$$8sin^{-1}(\frac{x}{4})-2x\frac{\sqrt{16-x^2}}{4}$$

$$8sin^{-1}(\frac{x}{4})-\frac{x}{2}\sqrt{16-x^2}$$

Your mistake was that your denominator should have been 4cosu, not $$4cos^2u$$

 

[Gib Z]

Your new so I won't say much, but please 1) Do not give out a full solution, it never helps the original poster learn anything and 2) Repeat what another member says just in a more blatantly obvious manner.

 

[flebbyman]

Sorry and thanks

 

[Gib Z]

I feel bad for being so harsh now :( Sorry lol.

 

[arildno]

I feel bad for being so harsh now :( Sorry lol.

I don't.
I'll flay him him next time.

Flebbyman: I know your adress..

 

[flebbyman]

I feel bad for being so harsh now :( Sorry lol.

Don't sweat it

I know your adress..