1. Integration problem

how to integrate (4-x^2)^(-3/2)?

2. Originally Posted by hkdrmark
how to integrate (4-x^2)^(-3/2)?
$\displaystyle \int (4-x^2)^{-3/2} ~dx$

Substitute:
$\displaystyle 2\sin(a) = x$
$\displaystyle 2\cos(a) ~da = dx$

$\displaystyle = 2\int cos(a) ~ (4-(2\sin(a))^2)^{-3/2} ~da$

$\displaystyle = 2\int cos(a) ~ (4-4\sin^2(a))^{-3/2} ~da$

$\displaystyle = 2\int cos(a) ~ 4^{-3/2} ~ (1-\sin^2(a))^{-3/2} ~da$

$\displaystyle = 2\int cos(a) ~ \frac 18 ~ (\cos^2(a))^{-3/2} ~da$

$\displaystyle = \frac 14 \int cos(a) ~ \sec^3(a) ~da$

$\displaystyle = \frac 14 \int \sec^2(a) ~da$

$\displaystyle = \frac 14 \tan(a) + C$

anti-substitute:
$\displaystyle a = \arcsin(x/2)$

$\displaystyle = \frac 14 \tan(\arcsin(x/2)) + C$

$\displaystyle = \frac x{4\sqrt{4-x^2}} + C$

3. thankyou

4. It's a classic integral, a reciprocal substitution also works.

5. Originally Posted by Krizalid
It's a classic integral, a reciprocal substitution also works.
Could you explain this? I (no joke) thought to myself "What would Krizalid do?" and so I tried making it [(2+x)(2-x)]^(-3/2) and substituting a=2+x, but that got me [a(4+a)]^(-3/2) which didn't seem to get any simpler to me, so I went with this method. I want to see what you're talking about though, I find your methods helpful.

6. Originally Posted by angel.white
Could you explain this? I (no joke) thought to myself "What would Krizalid do?" and so I tried making it [(2+x)(2-x)]^(-3/2) and substituting a=2+x, but that got me [a(4+a)]^(-3/2) which didn't seem to get any simpler to me, so I went with this method. I want to see what you're talking about though, I find your methods helpful.
Let $\displaystyle x = \frac{1}{u}$ so that $\displaystyle \frac{dx}{du} = -\frac{1}{u^2} \Rightarrow dx = - \frac{du}{u^2}$.

Then $\displaystyle I = - \int (4u^2 - 1)^{-3/2} \, u \, du$ and the solution easily follows.