Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 13

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