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tan^5x+tan^3x
tangent of to the power of 5x plus tangent of cubed x
tan5x+tan3x
tan⁵x+tan³x
tan to the power of 5x+tan to the power of 3x
tan^5x+tan^3xdx
Similar expressions
tan^5x-tan^3x

Solving integrals/ tan^5x+tan^3x

Integral of tan^5x+tan^3x dx

The solution

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  1                       
  /                       
 |                        
 |  /   5         3   \   
 |  \tan (x) + tan (x)/ dx
 |                        
/                         
0

$$\int\limits_{0}^{1} \left(\tan^{5}{\left(x \right)} + \tan^{3}{\left(x \right)}\right)\, dx$$

Integral(tan(x)^5 + tan(x)^3, (x, 0, 1))

Detail solution

Integrate term-by-term:
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 $\frac{du}{2}$ :
    
    $\int \frac{u^{2} - 2 u + 1}{4 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2} - 2 u + 1}{2 u}\, du = \frac{\int \frac{u^{2} - 2 u + 1}{u}\, du}{2}$
      
      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 $u = \sec^{4}{\left(x \right)}$ .
      
      Then let $du = 4 \tan{\left(x \right)} \sec^{4}{\left(x \right)} dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{4}$
      
      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 $u = \sec^{2}{\left(x \right)}$ .
      
      Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{2}$
      
      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 $- du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
      
      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 $u = \sec^{4}{\left(x \right)}$ .
      
      Then let $du = 4 \tan{\left(x \right)} \sec^{4}{\left(x \right)} dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{4}$
      
      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 $u = \sec^{2}{\left(x \right)}$ .
      
      Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{2}$
      
      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 $- du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
      
      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: $- \log{\left(\cos{\left(x \right)} \right)} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}$
1. Rewrite the integrand:
  
  $\tan^{3}{\left(x \right)} = \left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)}$
2. Let $u = \sec^{2}{\left(x \right)}$ .
  
  Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
  
  $\int \frac{u - 1}{4 u}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{u - 1}{2 u}\, du = \frac{\int \frac{u - 1}{u}\, du}{2}$
    1. Rewrite the integrand:
      
      $\frac{u - 1}{u} = 1 - \frac{1}{u}$
    2. Integrate term-by-term:
      1. The integral of a constant is the constant times the variable of integration:
        
        $\int 1\, du = u$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
        
        The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
        
        So, the result is: $- \log{\left(u \right)}$
      The result is: $u - \log{\left(u \right)}$
    So, the result is: $\frac{u}{2} - \frac{\log{\left(u \right)}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(\sec^{2}{\left(x \right)} \right)}}{2} + \frac{\sec^{2}{\left(x \right)}}{2}$
The result is: $\frac{\sec^{4}{\left(x \right)}}{4} - \frac{\sec^{2}{\left(x \right)}}{2}$
Now simplify:

$\frac{\left(\sec^{2}{\left(x \right)} - 2\right) \sec^{2}{\left(x \right)}}{4}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                              
 |                                 2         4   
 | /   5         3   \          sec (x)   sec (x)
 | \tan (x) + tan (x)/ dx = C - ------- + -------
 |                                 2         4   
/

$${{4\,\sin ^2x-3}\over{4\,\sin ^4x-8\,\sin ^2x+4}}-{{1}\over{2\, \sin ^2x-2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                          2   
1       1       -1 + 4*cos (1)
- + --------- - --------------
4        2             4      
    2*cos (1)     4*cos (1)

$${{\sin ^41}\over{4\,\sin ^41-8\,\sin ^21+4}}$$

                          2   
1       1       -1 + 4*cos (1)
- + --------- - --------------
4        2             4      
    2*cos (1)     4*cos (1)

$$- \frac{-1 + 4 \cos^{2}{\left(1 \right)}}{4 \cos^{4}{\left(1 \right)}} + \frac{1}{4} + \frac{1}{2 \cos^{2}{\left(1 \right)}}$$

Simplify

Numerical answer [src]

1.47078538753166

1.47078538753166

The graph

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

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Identical expressions

Similar expressions

Integral of tan^5x+tan^3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3