# Another Part where my brain's Fried

by Chaz706
Tags: brain, fried
 P: 13 $$\int \sqrt{4-x^2}^3 dx$$ I'm thinking integrating by parts would work, with u being that root and dv being dx, but is that the right method and direction. I've tried it and it seems more complicated.
 P: 12 Use $$x = 2 \sin \theta$$... then, $$dx = 2 \cos \theta d\theta$$, and your integral becomes: $$16 \int \cos^4 \theta d\theta$$ . Use $$\cos{2\theta} = 2\cos^2 \theta - 1$$, etc...
 P: 555 Damn it, no wonder I couldn't get it to work (I tried x = 2cos u)!
HW Helper
P: 11,722

## Another Part where my brain's Fried

So it's

$$\int \left(4-x^{2}\right)^{\frac{3}{2}} \ dx$$

How about a substitution

$$x=2\sin t$$ ?

Daniel.

EDIT:Didn't see the other posts.
 PF Patron Sci Advisor Thanks Emeritus P: 38,423 x= 2cos(u) should work exactly like "x= 2 sin(u)": dx= -2 sin u du and $$\sqrt{4- x^2}= \sqrt{4-4 cos^2(u)}= 2 sin(u)$$ so the integral becomes $$-16\int sin^4(u) du$$ and the only difference is that "-".
P: 555
 Quote by HallsofIvy x= 2cos(u) should work exactly like "x= 2 sin(u)": dx= -2 sin u du and $$\sqrt{4- x^2}= \sqrt{4-4 cos^2(u)}= 2 sin(u)$$ so the integral becomes $$-16\int sin^4(u) du$$ and the only difference is that "-".
Oops, I made a slight mistake!
 P: 13 Thanks once more! :)

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