<s>[INST] Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
 [/INST] $\bf{My\; solution::}$ Given $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
Let $$(\sec x+\tan x)= t\;,$$ Then $$\left(\sec x\cdot \tan x+\sec^2 x\right)dx = dt $$
So $$\displaystyle \left(\sec x+\tan x\right)dx = \frac{dt}{\sec x}\Rightarrow  \displaystyle dx = \frac{dt}{\sec x}$$
Now Using $$\bullet \displaystyle (\sec^2 x-\tan^2 x) = 1\Rightarrow (\sec x+\tan x)\cdot (\sec x-\tan x) = 1$$
So $$\displaystyle (\sec x-\tan x) = \frac{1}{t}$$ and we above substitute $$\displaystyle (\sec x+\tan x)=t$$
So we get $$\displaystyle \sec x = \frac{1}{2}\left(t+\frac{1}{t}\right)$$
Now our Integral convert into $$\displaystyle \frac{1}{2}\int\frac{\sec x}{t^{\frac{9}{2}}}dt = \frac{1}{2}\int \frac{t+\frac{1}{t}}{t^{\frac{9}{2}}}dt = \frac{1}{2}\int t^{-\frac{7}{2}}dt+\frac{1}{2}\int t^{-\frac{11}{2}}dt$$
So  we get $$\displaystyle \frac{1}{2}\cdot -\frac{2}{5}\cdot t^{-\frac{5}{2}}+\frac{1}{2}\cdot -\frac{2}{9}\cdot t^{-\frac{9}{2}}+\mathcal{C}$$
So $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = -\frac{1}{5}\cdot (\sec x+\tan x)^{-\frac{5}{2}}-\frac{1}{9}\cdot (\sec x+\tan x)^{-\frac{9}{2}}+\mathcal{C}$$
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