`int (sec^2x) / sqrt(25-tan^2x) dx` Find the indefinite integral

We have to evaluate the integral : `\int \frac{sec^2x}{\sqrt{25-tan^2x}}dx`

Let `tanx =t`

So, `sec^2x dx=dt`

Therefore we have,

`\int \frac{sec^2x}{\sqrt{25-tan^2x}}dx=\int \frac{dt}{\sqrt{25-t^2}}`

Now let `t=5sinu`

So, `dt= 5cosu du`

Hence we have,

`\int \frac{dt}{\sqrt{25-t^2}}=\int \frac{5cosu}{\sqrt{25-25sin^2u}}du`

               `=\int \frac{5cosu}{\sqrt{25(1-sin^2u)}}du`

                `=\int\frac{5cosu}{\sqrt{25cos^2u}}du `

                `=\int \frac{5cosu}{5cosu}du`

                 `=\int du`

                 `=u+C`   (where C is s constant)

                  `=\frac{1}{5}sin^{-1}(t)+C`

                  `=\frac{1}{5}sin^{-1}(tanx)+C`