Question

How do you evaluate this integral?

`int(x^(2))/((4-x^(2))^(3/2))dx`

Answer 1

The integral equals `x/sqrt(4 - x^2) - arcsin(x/2) + C`

Explanation:

Try using the trig substitution `x = 2sintheta`. Then `dx = 2costheta d theta` and so we can say

`I = int (2sintheta)^2/(4 - (2sintheta)^2)^(3/2) * 2costheta d theta`

`I = int (4sin^2theta)/(4 - 4sin^2theta)^(3/2) * 2costheta d theta`

`I = int (4sin^2theta)/(sqrt(4 - 4sin^2theta)^3) * 2costheta d theta`

`I = int (4sin^2theta)/sqrt((4cos^2theta)^3) * 2costheta d theta`

`I = int (4sin^2theta)/(8cos^3theta) * 2costheta d theta`

`I = int (sin^2theta)/cos^2thetad theta`

Which is equivalent to `tan^2theta`.

`I = tan^2theta d theta`

Use the identity `tan^2x = sec^2x - 1`.

`I = sec^2theta - 1 d theta`

This is a known integral

`I = tan theta - theta + C`

Reverse the substitutions. We know that `x/2 = sintheta`, and so `costheta = sqrt(4 - x^2)/2` and so `tantheta = (x/2)/(sqrt(4 - x^2)/2) = x/sqrt(4 - x^2)`

`I = x/sqrt(4 - x^2) - arcsin(x/2) +C`

Hopefully this helps!