[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.
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...
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.
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 "-".