Question

What is the integral of ${\tan ^5}\left( x \right)dx$?

Answer 1

Hint: To find the integral of ${\tan ^5}\left( x \right)dx$, first of all we have to write ${\tan ^5}\left( x \right)dx$ as ${\tan ^3}\left( x \right) \times {\tan ^2}\left( x \right)$. Then we have to use the identity $1 + {\tan ^2}x = {\sec ^2}x$ and substitute the value of ${\tan ^2}x$ in the obtained equation. After that simplify the equation and spate all the integration terms.
Now, we will have to use the property
$ \Rightarrow \int {{{\left[ {f\left( x \right)} \right]}^n} \cdot f'\left( x \right)dx = \dfrac{{{{\left[ {f\left( x \right)} \right]}^{n + 1}}}}{{n + 1}}} $ to find the integral.

Complete step by step solution:
In this question, we have to find the integral of ${\tan ^5}\left( x \right)dx$.
$ \Rightarrow I = \int {{{\tan }^5}\left( x \right)dx} $ - - - - - - - - - (1)
Now, we can write ${\tan ^5}\left( x \right)$ as ${\tan ^3}\left( x \right) \times {\tan ^2}\left( x \right)$. Therefore, equation (1) becomes,
$ \Rightarrow I = \int {{{\tan }^3}x \cdot {{\tan }^2}xdx} $- - - - - - - (2)
Now, we know that
$
   \Rightarrow 1 + {\tan ^2}x = {\sec ^2}x \\
   \Rightarrow {\tan ^2}x = {\sec ^2}x - 1 \\
 $
Putting the value of ${\tan ^2}x$ in equation (2), we get
$ \Rightarrow I = \int {{{\tan }^3}x \cdot \left( {{{\sec }^2}x - 1} \right)dx} $
$ \Rightarrow I = \int {{{\tan }^3}x \cdot {{\sec }^2}x - {{\tan }^3}xdx} $- - - - - - - (3)
Now, again we can write ${\tan ^3}x$ as ${\tan ^2}x \times \tan x$. Therefore, equation (3) becomes
$ \Rightarrow I = \int {{{\tan }^3}x \cdot {{\sec }^2}x - {{\tan }^2}x \cdot \tan xdx} $- - - - - (4)
Now, again substituting the value of ${\tan ^2}x$ in equation (4), we get
\[ \Rightarrow I = \int {\left( {\left( {{{\tan }^3}x \cdot {{\sec }^2}x} \right) - \left( {{{\sec }^2}x - 1} \right) \cdot \tan x} \right)dx} \]
\[ \Rightarrow I = \left( {\int {\left( {{{\tan }^3}x \cdot {{\sec }^2}x} \right) - \left( {{{\sec }^2}x \cdot \tan x} \right)} + \tan x} \right)dx\]- - - - - - - - (5)
Now, integrating each term separately in equation (5), we get
$ \Rightarrow I = \int {\left( {{{\tan }^3}x \cdot {{\sec }^2}x} \right)dx - \int {\left( {{{\sec }^2}x \cdot \tan x} \right)dx + \int {\tan xdx} } } $- - - - - (6)
Now, there is a proved theorem, that
$ \Rightarrow \int {{{\left[ {f\left( x \right)} \right]}^n} \cdot f'\left( x \right)dx = \dfrac{{{{\left[ {f\left( x \right)} \right]}^{n + 1}}}}{{n + 1}}} $
Here, in our equation (6) in first integration term,
$
   \Rightarrow f\left( x \right) = \tan x \\
   \Rightarrow n = 3 \\
   \Rightarrow f'\left( x \right) = {\sec ^2}x \\
 $
Hence,
\[ \Rightarrow \int {\left( {{{\tan }^3}x \cdot {{\sec }^2}x} \right)dx} = \dfrac{{{{\tan }^{3 + 1}}x}}{{3 + 1}} = \dfrac{{{{\tan }^4}x}}{4}\]
And in second integration term,
$
   \Rightarrow f\left( x \right) = \tan x \\
   \Rightarrow n = 1 \\
   \Rightarrow f'\left( x \right) = {\sec ^2}x \\
 $
Hence,
$ \Rightarrow \int {\left( {{{\sec }^2}x \cdot \tan x} \right)dx = \dfrac{{{{\tan }^{1 + 1}}x}}{{1 + 1}}} = \dfrac{{{{\tan }^2}x}}{2}$
And, $\int {\tan xdx = \ln \sec x} $
Therefore, equation (6) becomes
$ \Rightarrow I = \dfrac{{{{\tan }^4}x}}{4} - \dfrac{{{{\tan }^2}x}}{2} + \ln \sec x + c$
Where, c is the integration constant.
Hence, the integral of ${\tan ^5}\left( x \right)dx$ is $\dfrac{{{{\tan }^4}x}}{4} - \dfrac{{{{\tan }^2}x}}{2} + \ln \sec x + c$.

Note:
We cannot directly integrate trigonometric functions with power greater than 1 as there is no direct formula for it. We have to use the relations and formulas to simplify the expression so that we can integrate it easily. So, while finding the integral of trigonometric functions with power greater than 1, always look for the relations that can be used to simplify the expression.