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:
$\displaystyle a = \arcsin(x/2)$
$\displaystyle = \frac 14 \tan(\arcsin(x/2)) + C$
$\displaystyle = \frac x{4\sqrt{4-x^2}} + C$
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.