Mathematics - Indefinite Integration Question with Solution | TestHub

$\begin{array}{l}I=\int \frac{{\sec }^{2}x}{(\sec x+\tan x{)}^{9/2}}dx\\ \text{ Let }\sec x+\tan x=t\Rightarrow \sec x-\tan x=\frac{1}{t}\\ \Rightarrow \sec x=\frac{1}{2}\left(t+\frac{1}{t}\right)\text{ and }\sec x(\sec x+\tan x)dx=dt\\ \Rightarrow \sec xdx=\frac{dt}{t}\\ \therefore I=\frac{1}{2}\int \frac{\left(t+\frac{1}{t}\right)dt}{t^{9/2}\cdot t}=\frac{1}{2}\int \left(t^{-9/2}+t^{-13/2}\right)dt\\ =\frac{-1}{7}{t}^{-7/2}-\frac{1}{11}{t}^{-1/2}+K\\ =-\frac{1}{7t^{7/2}}-\frac{1}{11t^{11/2}}+K=-\frac{1}{t^{11/2}}\left(\frac{1}{11}+\frac{t^2}{7}\right)+K\\ =\frac{-1}{(\sec x+\tan x{)}^{11/2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\right\}+K\end{array}$