# What is the antiderivative of sec^5xtan^7x?

Mar 22, 2015

The answer is: ${\sec}^{11} \frac{x}{11} - {\sec}^{9} \frac{x}{3} + \frac{3 {\sec}^{7} x}{7} - {\sec}^{5} \frac{x}{5} + c$.

Remembering that:

• ${\sec}^{2} x = \frac{1}{\cos} ^ 2 x = \frac{{\sin}^{2} x + {\cos}^{2} x}{\cos} ^ 2 x =$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} \frac{x}{\cos} ^ 2 x = {\tan}^{2} x + 1 \Rightarrow$

${\tan}^{2} x = {\sec}^{2} x - 1$;

• $\int \sec x \tan x \mathrm{dx} = \sec x + c$;

• $\int {\left[f \left(x\right)\right]}^{n} \cdot f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c$.

Than:

$\int {\sec}^{5} x \cdot {\tan}^{7} x \mathrm{dx} = \int {\sec}^{5} x \cdot {\tan}^{6} x \cdot \tan x \mathrm{dx} =$

$= \int {\sec}^{5} x \cdot {\left({\sec}^{2} x - 1\right)}^{3} \cdot \tan x \mathrm{dx} =$

$= \int {\sec}^{5} x \left({\sec}^{6} x - 3 {\sec}^{4} x + 3 {\sec}^{2} x - 1\right) \cdot \tan x \mathrm{dx} =$

$= \int \left({\sec}^{11} x \tan x - 3 {\sec}^{9} x \tan x + 3 {\sec}^{7} x \tan x - {\sec}^{5} x \tan x\right) \mathrm{dx} =$

$= \int {\sec}^{10} x \sec x \tan x \mathrm{dx} - 3 \int {\sec}^{8} x \sec x \tan x \mathrm{dx} +$

$+ 3 \int {\sec}^{6} x \sec x \tan x \mathrm{dx} - \int {\sec}^{4} x \sec x \tan x \mathrm{dx} =$

$= {\sec}^{11} \frac{x}{11} - \frac{3 {\sec}^{9} x}{9} + \frac{3 {\sec}^{7} x}{7} - {\sec}^{5} \frac{x}{5} + c =$

$= {\sec}^{11} \frac{x}{11} - {\sec}^{9} \frac{x}{3} + \frac{3 {\sec}^{7} x}{7} - {\sec}^{5} \frac{x}{5} + c$.