The integral $\int \frac{{\sec }^{2}x}{(\sec x+\tan x{)}^{9/2}}dx$ equals (for some arbitrary constant K)

Solution:

We have,

$I=\int \frac{{\sec }^{2}x}{(\sec x+\tan x{)}^{9/2}}dx$

Let $secx+\tan x=t$.

. Then, $\sec x-\tan x=1/t$ and,

$\sec x(\sec x+\tan x)dx=dt$

$\therefore \mathrm{scc}xdx=\frac{1}{t}dt$ and, $\sec x=\frac{1}{2}\left(t+\frac{1}{t}\right)$

$\therefore I=\frac{1}{2}\int \frac{\frac{1}{t}\left(t+\frac{1}{t}\right)}{t^{9/2}}dt=\frac{1}{2}\int \frac{1}{t^{9/2}}+\frac{1}{t^{13/2}}dt$

$\begin{array}{l}\Rightarrow I=-\frac{1}{7t^{7/2}}-\frac{1}{11t^{11/2}}+k\\ \Rightarrow I=-\frac{1}{t^{11/2}}\left\{\frac{t^2}{7}+\frac{1}{11}\right\}+k\\ \Rightarrow I=-\frac{1}{(\sec x+\tan x{)}^{11/2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\right\}+k\end{array}$