Question

Integrals Giving Inverse Trigonometric Functions

∫ dx/√(9-x²)

Answer 1

We know that ,$\int \dfrac{dx}{\sqrt{a^2-x^2}}=sin^{-1}\dfrac{x}{a}+ C$

$\Rightarrow \int \dfrac{dx}{\sqrt{9-x^2}}=\int \dfrac{dx}{\sqrt{3^2-x^2}}= sin^{-1}\dfrac{x}{3} +C$