Solution assignment 02 Improper integrals

Calculate:

$\displaystyle\int_{0}^{a}{\frac{x}{\sqrt{a^2-x^2}}}dx$

The integrand is not bounded on the integration interval but it provides a finite result.

$\displaystyle\int_{0}^{a}{\frac{x}{\sqrt{a^2-x^2}}}dx=$

$=-\displaystyle\frac{1}{2}\int_{0}^{a}{\frac{d(a^2-x^2)}{\sqrt{a^2-x^2}}}=$

$=-\displaystyle\frac{1}{2}\int_{0}^{a}({a^2-x^2)^{-\frac{1}{2}}}d(a^2-x^2)=$

$=-\displaystyle\frac{1}{2}\cdot2[(a^2-x^2)^{\frac{1}{2}}]_{0}^{a}=a$