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

\(\begin{array}{l}

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

\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) d x=d t \\

\Rightarrow \sec x d x=\frac{d t}{t} \\

\therefore \quad I=\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right) d t}{t^{9 / 2} \cdot t}=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t \\

=\frac{-1}{7} t^{-7 / 2}-\frac{1}{11} t^{-1 / 2}+K \\

=-\frac{1}{7 t^{7 / 2}}-\frac{1}{11 t^{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}\)