Chapter 5.7, Problem 8E

To determine

To calculate: the integral value of ${\displaystyle \int {\tan }^{5}x{\sec }^{3}xdx}$

Answer to Problem 8E

The integration of ${\displaystyle \int {\tan }^{5}x{\sec }^{3}xdx}$ is $\frac{1}{7}{\sec }^{7}x-\frac{2}{5}{\sec }^{5}x+\frac{1}{3}{\sec }^{3}x+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int {\tan }^{5}x{\sec }^{3}xdx}$

Calculation: ${\displaystyle \int {\tan }^{5}x{\sec }^{3}xdx}$

Rearranging and using trigonometric identity

  ${\displaystyle \int {\tan }^{4}x{\sec }^{2}x\tan x\sec xdx}={\displaystyle \int {\left({\sec }^{2}x-1\right)}^2{\sec }^{2x}\sec x\tan xdx}$

Substituting $u=\sec x$ the new integral will be $du=\sec x\tan xdx$

  $\begin{array}{l}={\displaystyle \int {\left(u^2-1\right)}^2{u}^{2}du}\\ ={\displaystyle \int \left(u^4-2u^2+1\right)u^{2}du}\\ ={\displaystyle \int \left(u^6-2u^4+u^2\right)du}\\ =\frac{u^7}{7}+\frac{2u^5}{5}+\frac{u^3}{3}+C\\ =\frac{1}{7}{\sec }^{7}x-\frac{2}{5}{\sec }^{5}x+\frac{1}{3}{\sec }^{3}x+C\end{array}$

Hence,

The integration of ${\displaystyle \int {\tan }^{5}x{\sec }^{3}xdx}$ is $\frac{1}{7}{\sec }^{7}x-\frac{2}{5}{\sec }^{5}x+\frac{1}{3}{\sec }^{3}x+C$