Evaluate: `int(dx)/((a^2-x^2)^(3/2))`

Explanation

Put `x=a tan theta implies dx = a sec^(2) theta d theta`
` :. I= int (dx)/(a^(2)+x^(2))^(3//2)`
`= int(a sec^(2)theta)/((a^(2)+a^(2) tan^(2)theta)^(3//2))d theta`
`=int(a sec^(2)theta)/(a^(3)(sec^(2)theta)^(3//2))d theta`
`=(1)/(a^(2))int(d theta)/(sec theta)`
`=(1)/(a^(2))int cos theta d theta`
`=(1)/(a^(2))sin theta +c`
Now `tan theta=(x)/(a)`
` :. sin theta = (x)/(sqrt(x^(2)+a^(2)))`
` :. I= (x)/(a^(2)sqrt(x^(2)+a^(2)))+c`