Question:

Let $t = \sec x + \tan x$. Then $dt = \sec x (\sec x + \tan x) dx = t \sec x dx \implies \sec x dx = \frac{dt}{t}$.<br>We know $\sec x - \tan x = \frac{1}{\sec x + \tan x} = \frac{1}{t}$.<br>Adding $\sec x + \tan x = t$ and $\sec x - \tan x = \frac{1}{t}$, we get $2 \sec x = t + \frac{1}{t} \implies \sec x = \frac{1}{2}(t + \frac{1}{t})$.<br>The integral becomes $I = \int \frac{\sec x \cdot \sec x dx}{(\sec x + \tan x)^{9/2}} = \int \frac{\frac{1}{2}(t + t^{-1}) \frac{dt}{t}}{t^{9/2}} = \frac{1}{2} \int \frac{1 + t^{-2}}{t^{9/2}} dt = \frac{1}{2} \int (t^{-9/2} + t^{-13/2}) dt$.<br>$I = \frac{1}{2} \left[ \frac{t^{-7/2}}{-7/2} + \frac{t^{-11/2}}{-11/2} \right] + K = -\frac{1}{7} t^{-7/2} - \frac{1}{11} t^{-11/2} + K$.<br>$I = -\frac{1}{t^{11/2}} \left[ \frac{1}{11} + \frac{1}{7} t^2 \right] + K = -\frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7} (\sec x + \tan x)^2 \right\} + K$.