# Integral of Sec^2 x/(secax+tanx) ^9/2 dx=?

Mar 19, 2018

$- \frac{1}{7} {\left(\sec x + \tan x\right)}^{- \frac{7}{2}} - \frac{1}{11} {\left(\sec x + \tan x\right)}^{- \frac{11}{2}} + C$

#### Explanation:

$\int {\left(\sec x\right)}^{2} / {\left(\sec x + \tan x\right)}^{\frac{9}{2}} \cdot \mathrm{dx}$

=$\frac{1}{2} \int \frac{2 {\left(\sec x\right)}^{2}}{\sec x + \tan x} ^ \left(\frac{9}{2}\right) \cdot \mathrm{dx}$

=$\frac{1}{2} \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x} ^ \left(\frac{9}{2}\right) \mathrm{dx}$+$\frac{1}{2} \int \sec x \cdot \frac{\sec x - \tan x}{\sec x + \tan x} ^ \left(\frac{9}{2}\right) \mathrm{dx}$

=$\frac{1}{2} \int \frac{{\left(\sec x\right)}^{2} + \sec x \cdot \tan x}{\sec x + \tan x} ^ \left(\frac{9}{2}\right) \mathrm{dx}$+$\frac{1}{2} \int \sec x \cdot \frac{{\left(\sec x\right)}^{2} - {\left(\tan x\right)}^{2}}{\sec x + \tan x} ^ \left(\frac{11}{2}\right) \mathrm{dx}$

=$\frac{1}{2} \int \frac{{\left(\sec x\right)}^{2} + \sec x \cdot \tan x}{\sec x + \tan x} ^ \left(\frac{9}{2}\right) \mathrm{dx}$+$\frac{1}{2} \int \sec \frac{x}{\sec x + \tan x} ^ \left(\frac{11}{2}\right) \mathrm{dx}$

=$\frac{1}{2} \int \frac{{\left(\sec x\right)}^{2} + \sec x \cdot \tan x}{\sec x + \tan x} ^ \left(\frac{9}{2}\right) \mathrm{dx}$+$\frac{1}{2} \int \frac{{\left(\sec x\right)}^{2} + \sec x \cdot \tan x}{\sec x + \tan x} ^ \left(\frac{13}{2}\right) \mathrm{dx}$

After using $y = \sec x + \tan x$ and $\mathrm{dy} = \left[{\left(\sec x\right)}^{2} + \sec x \cdot \tan x\right] \cdot \mathrm{dx}$ transforms, this integral became

$\frac{1}{2} \int \frac{\mathrm{dy}}{y} ^ \left(\frac{9}{2}\right)$+$\frac{1}{2} \int \frac{\mathrm{dy}}{y} ^ \left(\frac{13}{2}\right) \mathrm{dx}$

=$- \frac{1}{7} {y}^{- \frac{7}{2}} - \frac{1}{11} {y}^{- \frac{11}{2}} + C$

=$- \frac{1}{7} {\left(\sec x + \tan x\right)}^{- \frac{7}{2}} - \frac{1}{11} {\left(\sec x + \tan x\right)}^{- \frac{11}{2}} + C$