Question

Evaluate the integral\(\int {{{\tan }^4}} x{\sec ^6}xdx\)

Short answer

The value of \(\int {{{\tan }^4}} x{\sec ^6}xdx\) is \(\frac{{{{\tan }^5}x}}{5} + \frac{{{{\tan }^7}x}}{7} + c\)

Step-by-step solution

Given Data

Given data is\(\int {{{\tan }^4}} x{\sec ^6}xdx\)

Evaluating the integration:

\(\int {{{\tan }^4}} x{\sec ^6}xdx = \int {({{\tan }^4}x)(1 + {{\tan }^2}} x)({\sec ^2}x)dx\)

From the Trigonometric Identities, we can write\({\sec ^2}x = 1 + {\tan ^2}x\)

=\(\int {{{\tan }^4}} x{\sec ^6}xdx = \int {({{\tan }^4}} x + {\tan ^6}x){\sec ^2}xdx\)

=\(\int {{{\tan }^4}} x{\sec ^6}xdx = \int {({{\tan }^4}x)} ({\sec ^2}x)dx + \int {({{\tan }^6}} x)({\sec ^2}x)dx\)à1

Finding the equation 1 value:

Let\(u = \tan x\)

Differentiating with respect to\(x\)

\(\begin{aligned}{l}\frac{{du}}{{dx}} = {\sec ^2}x\\ = du = {\sec ^2}xdx\end{aligned}\)

Substituting the above value in equation one

\(\begin{aligned}{l}\int {{{\tan }^4}} x{\sec ^6}xdx &= \int {{u^4}} .du + \int {{u^6}.du} \\\int {{{\tan }^4}} x{\sec ^6}xdx &= \frac{{{u^5}}}{5} + \frac{{{u^7}}}{7} + c\end{aligned}\)à 2

Substituting the original values in equation 2

Since\(u = \tan x\), therefore equation 2 becomes

\(\begin{aligned}{l}\int {{{\tan }^4}} x{\sec ^6}xdx &= \frac{{{{(\tan x)}^5}}}{5} + \frac{{{{(\tan x)}^7}}}{7} + c\\\int {{{\tan }^4}} x{\sec ^6}xdx &= \frac{{{{\tan }^5}x}}{5} + \frac{{{{\tan }^7}x}}{7} + c\end{aligned}\)

Hence the answer for the given equation is

\(\begin{aligned}{l}\\\int {{{\tan }^4}} x{\sec ^6}xdx &= \frac{{{{\tan }^5}x}}{5} + \frac{{{{\tan }^7}x}}{7} + c\end{aligned}\)