Evaluate $\int \frac{\sec^{2}{x}}{(\sec{x} + \tan{x})^{5/2}} \ \mathrm dx$

Question

I am unable to solve the following integral: $$\int \frac{\sec^{2}{x}}{(\sec{x} + \tan{x})^{5/2}} \ \mathrm dx.$$

Tried putting $t=\tan{x}$ so that numerator has $\sec^{2}{x}$ but that doesn't help.

Answer 1

Here are some hints:

• Put $t = \sec{x} + \tan{x}$.

• Use the identity $\sec^{2}{x} - \tan^{2}{x} = 1$, to get the value of $\sec{x}$ in terms of $t$.

Complete Solution:

$t = \sec{x} + \tan{x} \Rightarrow dt = \sec{x} \cdot (\sec{x}+\tan{x}) dx \Rightarrow dt = \Bigl(t+\frac{1}{t}\Bigr)\cdot t \ dx$. By following the above you get $t+\frac{1}{t} = \sec{x}$. So your integral becomes \begin{align*} \int \frac{\sec^{2}{x}}{(\sec{x}+\tan{x})^{5/2}}\ dx &= \int \frac{1}{t^{5/2}} \times \biggl(t+\frac{1}{t}\biggr)^{2} \times \frac{dt}{(t+\frac{1}{t}) \cdot t}\end{align*}