`int dx/sqrt(9-x^2)` Find the indefinite integral

We have to evaluate the integral: `\int \frac{dx}{\sqrt{9-x^2}}`

let `x=3sint`

So, `dx=3cost dt`

Hence we have,

`\int \frac{dx}{\sqrt{9-x^2}}=\int \frac{3cost}{\sqrt{9-9sin^2t}}dt`

              `=\int \frac{3cost}{\sqrt{9(1-sin^2t)}} dt`

               `=\int \frac{3cost}{\sqrt{9cos^2t}}dt`

                `=\int \frac{3cost}{3cost}dt`

                 `=\int dt`

                  `=t+C` (where C is a constant)

                  `=\sin^{-1}(x/3)+C`