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Identical expressions
4tan^ five (x)
4 tangent of to the power of 5(x)
4 tangent of to the power of five (x)
4tan5(x)
4tan5x
4tan⁵(x)
4tan^5x
4tan^5(x)dx

Integral of 4tan^5(x) dx

The solution

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 pi             
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 4              
  /             
 |              
 |       5      
 |  4*tan (x) dx
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/               
0

$$\int\limits_{0}^{\frac{\pi}{4}} 4 \tan^{5}{\left(x \right)}\, dx$$

Integral(4*tan(x)^5, (x, 0, pi/4))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int 4 \tan^{5}{\left(x \right)}\, dx = 4 \int \tan^{5}{\left(x \right)}\, dx$
1. Rewrite the integrand:
  
  $\tan^{5}{\left(x \right)} = \left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)}$
2. There are multiple ways to do this integral.
  Method #1
  1. Let $u = \sec^{2}{\left(x \right)}$ .
    
    Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u^{2} - 2 u + 1}{2 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2} - 2 u + 1}{u}\, du = \frac{\int \frac{u^{2} - 2 u + 1}{u}\, du}{2}$
      
      Rewrite the integrand:
      
      $\frac{u^{2} - 2 u + 1}{u} = u - 2 + \frac{1}{u}$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-2\right)\, du = - 2 u$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      The result is: $\frac{u^{2}}{2} - 2 u + \log{\left(u \right)}$
      
      So, the result is: $\frac{u^{2}}{4} - u + \frac{\log{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(\sec^{2}{\left(x \right)} \right)}}{2} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}$
  Method #2
  1. Rewrite the integrand:
    
    $\left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)} = \tan{\left(x \right)} \sec^{4}{\left(x \right)} - 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} + \tan{\left(x \right)}$
  2. Integrate term-by-term:
    1. Let $u = \sec{\left(x \right)}$ .
      
      Then let $du = \tan{\left(x \right)} \sec{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{3}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{4}{\left(x \right)}}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\, dx = - 2 \int \tan{\left(x \right)} \sec^{2}{\left(x \right)}\, dx$
      
      Let $u = \sec{\left(x \right)}$ .
      
      Then let $du = \tan{\left(x \right)} \sec{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{2}{\left(x \right)}}{2}$
      
      So, the result is: $- \sec^{2}{\left(x \right)}$
    1. Rewrite the integrand:
      
      $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
    2. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(\cos{\left(x \right)} \right)}$
    The result is: $- \log{\left(\cos{\left(x \right)} \right)} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}$
  Method #3
  1. Rewrite the integrand:
    
    $\left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)} = \tan{\left(x \right)} \sec^{4}{\left(x \right)} - 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} + \tan{\left(x \right)}$
  2. Integrate term-by-term:
    1. Let $u = \sec{\left(x \right)}$ .
      
      Then let $du = \tan{\left(x \right)} \sec{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{3}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{4}{\left(x \right)}}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\, dx = - 2 \int \tan{\left(x \right)} \sec^{2}{\left(x \right)}\, dx$
      
      Let $u = \sec{\left(x \right)}$ .
      
      Then let $du = \tan{\left(x \right)} \sec{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{2}{\left(x \right)}}{2}$
      
      So, the result is: $- \sec^{2}{\left(x \right)}$
    1. Rewrite the integrand:
      
      $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
    2. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(\cos{\left(x \right)} \right)}$
    The result is: $- \log{\left(\cos{\left(x \right)} \right)} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}$
So, the result is: $2 \log{\left(\sec^{2}{\left(x \right)} \right)} + \sec^{4}{\left(x \right)} - 4 \sec^{2}{\left(x \right)}$
Add the constant of integration:

$2 \log{\left(\sec^{2}{\left(x \right)} \right)} + \sec^{4}{\left(x \right)} - 4 \sec^{2}{\left(x \right)}+ \mathrm{constant}$

The answer is:

$2 \log{\left(\sec^{2}{\left(x \right)} \right)} + \sec^{4}{\left(x \right)} - 4 \sec^{2}{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                       
 |                                                        
 |      5                4           2           /   2   \
 | 4*tan (x) dx = C + sec (x) - 4*sec (x) + 2*log\sec (x)/
 |                                                        
/

$$\int 4 \tan^{5}{\left(x \right)}\, dx = C + 2 \log{\left(\sec^{2}{\left(x \right)} \right)} + \sec^{4}{\left(x \right)} - 4 \sec^{2}{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

          /  ___\
          |\/ 2 |
-1 - 4*log|-----|
          \  2  /

$$-1 - 4 \log{\left(\frac{\sqrt{2}}{2} \right)}$$

          /  ___\
          |\/ 2 |
-1 - 4*log|-----|
          \  2  /

$$-1 - 4 \log{\left(\frac{\sqrt{2}}{2} \right)}$$

-1 - 4*log(sqrt(2)/2)

Numerical answer [src]

0.386294361119891

0.386294361119891

Use the examples entering the upper and lower limits of integration.

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How to use it?

Identical expressions

Integral of 4tan^5(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3