Calculus Problem

[iggybaseball]

I am having trouble solving this integral:

\frac{\1}{(4-z^2)^(3/2)}

I know that x = asin(theta)
theta = arcsin (x/a)
d(theta) = 1 / sqrt(4-z^2) dz

but then I get stuck. Could someone give me a hand?

ps there should be a number 1 on top of the fraction and the integral has dz after it respectfully. I couldn't get these two in ( It is the first time I used latex) Thanks

[George Jones]

Choose z = asin(theta). With regards to your problem, what should you choose a to be? Differentiate z = asin(theta) to find dz = ?

Regards,
George

[HallsofIvy]

It's the same thing but simpler: let z= 2sin(θ) then dz= 2cos(θ).
\frac{1}{(4-z^)^{\frac{3}{2}}}= \frac{1}{8(1- sin^2(\theta))^{\frac{3}{2}}}= \frac{1}{8 cos^3(\theta)}
The integral becomes \frac{1}{8}\integral{\frac{d\theta}{cos^2(\theta)} .

The way you were doing it works too, of course.
Since d\theta= \frac{dz}{\sqrt{((4-z^2)}} and \frac{1}{(4-z^2)^{\frac{3}{2}}}= \frac{1}{4-z^2}\frac{1}{\sqrt{4-z^2}}, that gives the same thing.