Question

Integrals Giving Inverse Trigonometric Functions

∫(xdx)/√(4-x²)

Answer 1

$\int{\frac{x}{\sqrt{4-x^2}}dx}$

$\ let\ u=\ \sqrt{4-x^2}$

$\ du=\ -\frac{x}{\sqrt{4-x^2}}dx$

$\ \int{-du\ }$

$-u + c$

$-\ \sqrt{4-x^2}+c$