What U substitution is being used here? (direct integral calculation of charge)

Question

In my physics reading, we have the integral over $x_s$ from -L/2 to L/2 of $\frac{1}{((y^2+x_s^2)^{3/2})}$. Somehow, from this, the reading gets $\frac{x_s}{y^2*\sqrt{y^2+x_s^2}}$ from -L/2 to L/2. I'm assuming this was done via u-substitution, but every u value I've tried hasn't worked. How did they get from the initial integral to the result? Thank you for your help.

Answer 1

This integral is best handled by trigonometric substitution of the form $x_s=y\tan(\theta)$, $dx_s=y\sec^2(\theta) d\theta$ or $x_s=y\sinh(\phi)$, $dx_s=y\cosh(\phi)d\phi$. The resulting integrals after these substitutions are simple trigonometric integrals in $\theta$ or $\phi$ and can readily be evaluated. In this case, instead of mapping the bounds to $\theta$ or $\phi$, back substitution was used to cast the antiderivative into terms of $x_s$.

In general, whenever the integrand is of the form $g(a^2+[f(x)]^2)$ for functions $f,g$ and constant $a$, it may be helpful to make the substitution $f(x)=a\tan(\theta)$ or $f(x)=a\sinh(\phi)$. This is done to exploit the Pythagorean identity in the form $1+\tan^2(\theta)=\sec^2(\theta)$ or the corresponding hyperbolic identity in the form $1+\sinh^2(\phi)=\cosh^2(\phi)$. This particular problem has $f(x)=x$, $g(x)=x^{-3/2}$, and $a=y$.

If you are set on using $u$-substitution, the clever substitution $u=\frac{y^2}{x_s^2}+1$ suggested by @Quanto works!