Question

Evaluate the value of $\int {{{\tan }^4}xdx} $ .

Answer 1

Hint:
Here, we are asked to find the value of $\int {{{\tan }^4}xdx} $ .
Let, $I = \int {{{\tan }^4}xdx} $ and write ${\tan ^4}x$ as ${\tan ^2}x \cdot {\tan ^2}x$ .
Then, put ${\tan ^2}x = {\sec ^2}x - 1$ in any one of the values.
Thus, solve the value of I further to get the required answer.

Complete step by step solution:
Here, we are asked to find the value of $\int {{{\tan }^4}xdx} $ .
Let, $I = \int {{{\tan }^4}xdx} $ .
Now, we can write ${\tan ^4}x$ as ${\tan ^2}x \cdot {\tan ^2}x$
 $\therefore I = \int {{{\tan }^2}x \cdot {{\tan }^2}xdx} $
Also, we know that ${\tan ^2}x = {\sec ^2}x - 1$
 $\therefore I = \int {{{\tan }^2}x\left( {{{\sec }^2}x - 1} \right)dx} $
 $\therefore I = \int {\left( {{{\tan }^2}x{{\sec }^2}x - {{\tan }^2}x} \right)dx} $
 $
  \therefore I = \int {{{\tan }^2}x{{\sec }^2}xdx} - \int {{{\tan }^2}xdx} \\
  \therefore I = \int {{{\tan }^2}x{{\sec }^2}xdx} - \int {\left( {{{\sec }^2}x - 1} \right)dx} \\
  \therefore I = \int {{{\tan }^2}x{{\sec }^2}xdx} - \int {{{\sec }^2}xdx} + \int {dx} \\
  \therefore I = \int {{{\tan }^2}x{{\sec }^2}xdx} - \int {{{\sec }^2}xdx} + x + C \\
 $
Now, to solve further, let $\tan x = t$
 $\therefore {\sec ^2}xdx = dt$
 $\therefore I = \int {{t^2}dt} - \int {dt} + x + C$
 $\therefore I = \dfrac{{{t^3}}}{3} - t + x + C$
Now, we shall return back the substituted value $\tan x = t$ in the above equation.
 $\therefore I = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C$

Thus, $\int {{{\tan }^4}xdx} = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C$.


Note:
Alternate method:
Here, we are asked to find the value of $\int {{{\tan }^4}xdx} $ .
Let, $I = \int {{{\tan }^4}xdx} $ .
Now, $I = \int {\left( {{{\tan }^4}x + 1 - 1} \right)dx} $
 $
  \therefore I = \int {\left( {{{\tan }^4}x - 1} \right)dx} + \int {1dx} \\
  \therefore I = \int {\left( {{{\tan }^2}x - 1} \right)\left( {{{\tan }^2}x + 1} \right)dx} + x + C \\
 $
Since, ${\tan ^2}x + 1 = {\sec ^2}x$
 $\therefore I = \int {{{\sec }^2}x\left( {{{\tan }^2}x - 1} \right)dx} + x + C$
 $\therefore I = \int {{{\tan }^2}x{{\sec }^2}xdx} - \int {{{\sec }^2}xdx} + x + C$
Now, to solve further, let $\tan x = t$
 $\therefore {\sec ^2}xdx = dt$
 $\therefore I = \int {{t^2}dt} - \int {dt} + x + C$
 $\therefore I = \dfrac{{{t^3}}}{3} - t + x + C$
Now, we shall return back the substituted value $\tan x = t$ in the above equation.
 $\therefore I = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C$
Thus, $\int {{{\tan }^4}xdx} = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C$ .