Integral of (4-x^2)^(-3/2)

I tried a trig sub, but I must have done it wrong, because my answer isn't even close to the book's.

Question

Integral of (4-x^2)^(-3/2)

I tried a trig sub, but I must have done it wrong, because my answer isn't even close to the book's.

mathmath333 · Answer

which sub u tried

theoreo · Answer

I tried $$x = 2\sin(\theta)$$ 

which gave me $$4-4\sin^2 = 4(1-\sin^2) =( 4*\cos^2(\theta))^-3/2$$

mathmath333 · Answer

this sub should work perfectly

theoreo · Answer

Ok. So if $$(\cos^2)^-3/2 = (\sqrt{\cos^2})^-3= \cos^-3$$ right? I got stuck trying to integrate $$\sec^3$$ and the answer in my book doesn't involve trig.

mathmath333 · Answer

one cos term should be cancelled and u should have $\dfrac{\sec^2t}{4}$ right?

mathmath333 · Answer

i mean $\dfrac{\sec^2\theta}{4}$

theoreo · Answer

I don't see how a term would cancel. Could you explain?

mathmath333 · Answer

u have to change $dx$ to $d\theta$ so

$x = 2\sin(\theta)\
\dfrac{dx}{d\theta}=2\cos(\theta)$   


had u added that while calculation

theoreo · Answer

Ah I forgot about that part. Thanks a lot.