Question

How do you evaluate the integral `int dx/(1-x^2)^(5/2)`?

Answer 1

`(x(3-2x^2))/(3(1-x^2)^(3/2))+C`

Explanation:

Let `x=sintheta`. This implies that `dx=costhetad theta`.

`intdx/(1-x^2)^(5/2)`

`=int(costhetad theta)/(1-sin^2theta)^(5/2)`

`=intcostheta/(cos^2theta)^(5/2)d theta`

`=intsec^4thetad theta`

`=intsec^2theta(1+tan^2theta)d theta`

Let `u=tantheta` so `du=sec^2thetad theta`.

`=int(1+u^2)du`

`=u+1/3u^3+C`

`=tantheta+1/3tan^3theta+C`

`=1/3tantheta(3+tan^2theta)+C`

Where `x=sintheta`, we have a triangle where the opposite side is `x` and the hypotenuse is `1`. Thus, the adjacent side is `sqrt(1-x^2)` and `tantheta=x//sqrt(1-x^2)`.

`=x/(3sqrt(1-x^2))(3+x^2/(1-x^2))`

`=1/(3sqrt(1-x^2))((3-x^2)/(1-x^2))`

`=(x(3-2x^2))/(3(1-x^2)^(3/2))+C`