Another Part where my brain's Fried

[Chaz706]

$$\int \sqrt{4-x^2}^3 dx$$

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.

 

[z0r]

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...

 

[Nylex]

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

 

[dextercioby]

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.

 

[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 "-".

 

[Nylex]

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!

 

[Chaz706]

Thanks once more! :)