`int dx/sqrt(1-4x^2)` Find the indefinite integral

We have to evaluate `\int \frac{dx}{\sqrt{1-4x^2}}`

Let `x=\frac{1}{2} sint `

So, `dx= \frac{1}{2}cost dt`

Hence we have,

`\int \frac{dx}{\sqrt{1-4x^2}}=\int \frac{\frac{1}{2}cost }{\sqrt{1-4(\frac{1}{4}sin^2t)}}dt`

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

`= int 1/2cost/sqrt(cos^2t) dt`

                `=\int \frac{\frac{1}{2}cost}{cost}dt`

                 `=\int \frac{1}{2}dt`

                 `=\frac{1}{2}t+C`

                  `=\frac{1}{2}sin^{-1}(2x)+C`