Question

Evaluate:
$\underset{-a}{\overset{a}{\int }}\frac{dy}{(x^2+y^2{)}^{\frac{3}{2}}}$

Answer 1

Apply substitution,

$y=xtan\left(u\right)=\int \frac{1}{x^2\sec \left(u\right)}du=\frac{1}{x^2}\cdot \int \cos \left(u\right)du=\frac{1}{x^2}\sin \left(u\right)$

now,substitute back

$=\frac{1}{x^2}\sin \left({\tan }^{-1}\left(\frac{1}{x}y\right)\right)=\frac{y}{x^2\sqrt{x^2+y^2}}$

Computing the boundary points,

${\int }_{-a}^a\frac{1}{{(x^2+y^2)}^{\frac{3}{2}}}dy=\frac{a}{x^2\sqrt{x^2+a^2}}-(-\frac{a}{x^2\sqrt{x^2+a^2}})=\frac{a}{x^2\sqrt{x^2+a^2}}-(-\frac{a}{x^2\sqrt{x^2+a^2}})=\frac{2a}{x^2\sqrt{x^2+a^2}}$