Another Part where my brain's Fried

Chaz706

[tex] \int \sqrt{4-x^2}^3 dx [/tex]

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.

z0r

Use [tex]x = 2 \sin \theta [/tex]...
then,
[tex]dx = 2 \cos \theta d\theta [/tex],
and your integral becomes:
[tex]16 \int \cos^4 \theta d\theta [/tex] .
Use [tex]\cos{2\theta} = 2\cos^2 \theta - 1 [/tex], etc...

Nylex

Damn it, no wonder I couldn't get it to work (I tried x = 2cos u)!

dextercioby

So it's

[tex]\int \left(4-x^{2}\right)^{\frac{3}{2}} \ dx[/tex]

How about a substitution

[tex] x=2\sin t [/tex] ?

Daniel.

EDIT:Didn't see the other posts.

HallsofIvy

x= 2cos(u) should work exactly like "x= 2 sin(u)": dx= -2 sin u du and
[tex]\sqrt{4- x^2}= \sqrt{4-4 cos^2(u)}= 2 sin(u)[/tex] so the integral becomes
[tex]-16\int sin^4(u) du[/tex] and the only difference is that "-".

Nylex

Oops, I made a slight mistake!

Chaz706

Thanks once more! :)