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