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Question

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

What is the integral of tan5(x)dx{\tan ^5}\left( x \right)dx?

Explanation

Solution

To find the integral of tan5(x)dx{\tan ^5}\left( x \right)dx, first of all we have to write tan5(x)dx{\tan ^5}\left( x \right)dx as tan3(x)×tan2(x){\tan ^3}\left( x \right) \times {\tan ^2}\left( x \right). Then we have to use the identity 1+tan2x=sec2x1 + {\tan ^2}x = {\sec ^2}x and substitute the value of tan2x{\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
[f(x)]nf(x)dx=[f(x)]n+1n+1\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 tan5(x)dx{\tan ^5}\left( x \right)dx.
I=tan5(x)dx\Rightarrow I = \int {{{\tan }^5}\left( x \right)dx} - - - - - - - - - (1)
Now, we can write tan5(x){\tan ^5}\left( x \right) as tan3(x)×tan2(x){\tan ^3}\left( x \right) \times {\tan ^2}\left( x \right). Therefore, equation (1) becomes,
I=tan3xtan2xdx\Rightarrow I = \int {{{\tan }^3}x \cdot {{\tan }^2}xdx}- - - - - - - (2)
Now, we know that
1+tan2x=sec2x tan2x=sec2x1  \Rightarrow 1 + {\tan ^2}x = {\sec ^2}x \\\ \Rightarrow {\tan ^2}x = {\sec ^2}x - 1 \\\
Putting the value of tan2x{\tan ^2}x in equation (2), we get
I=tan3x(sec2x1)dx\Rightarrow I = \int {{{\tan }^3}x \cdot \left( {{{\sec }^2}x - 1} \right)dx}
I=tan3xsec2xtan3xdx\Rightarrow I = \int {{{\tan }^3}x \cdot {{\sec }^2}x - {{\tan }^3}xdx}- - - - - - - (3)
Now, again we can write tan3x{\tan ^3}x as tan2x×tanx{\tan ^2}x \times \tan x. Therefore, equation (3) becomes
I=tan3xsec2xtan2xtanxdx\Rightarrow I = \int {{{\tan }^3}x \cdot {{\sec }^2}x - {{\tan }^2}x \cdot \tan xdx}- - - - - (4)
Now, again substituting the value of tan2x{\tan ^2}x in equation (4), we get
I=((tan3xsec2x)(sec2x1)tanx)dx\Rightarrow I = \int {\left( {\left( {{{\tan }^3}x \cdot {{\sec }^2}x} \right) - \left( {{{\sec }^2}x - 1} \right) \cdot \tan x} \right)dx}
I=((tan3xsec2x)(sec2xtanx)+tanx)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
I=(tan3xsec2x)dx(sec2xtanx)dx+tanxdx\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
[f(x)]nf(x)dx=[f(x)]n+1n+1\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,
f(x)=tanx n=3 f(x)=sec2x  \Rightarrow f\left( x \right) = \tan x \\\ \Rightarrow n = 3 \\\ \Rightarrow f'\left( x \right) = {\sec ^2}x \\\
Hence,
(tan3xsec2x)dx=tan3+1x3+1=tan4x4\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,
f(x)=tanx n=1 f(x)=sec2x  \Rightarrow f\left( x \right) = \tan x \\\ \Rightarrow n = 1 \\\ \Rightarrow f'\left( x \right) = {\sec ^2}x \\\
Hence,
(sec2xtanx)dx=tan1+1x1+1=tan2x2\Rightarrow \int {\left( {{{\sec }^2}x \cdot \tan x} \right)dx = \dfrac{{{{\tan }^{1 + 1}}x}}{{1 + 1}}} = \dfrac{{{{\tan }^2}x}}{2}
And, tanxdx=lnsecx\int {\tan xdx = \ln \sec x}
Therefore, equation (6) becomes
I=tan4x4tan2x2+lnsecx+c\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 tan5(x)dx{\tan ^5}\left( x \right)dx is tan4x4tan2x2+lnsecx+c\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.