Integral of tan^5xdx dx

The solution

You have entered [src]

  1             
  /             
 |              
 |     5        
 |  tan (x)*1 dx
 |              
/               
0

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

Integral(tan(x)^5*1, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\tan^{5}{\left(x \right)} 1 = \left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)}$
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:
    
    $\int \frac{u^{2} - 2 u + 1}{2 u}\, du = \frac{\int \frac{u^{2} - 2 u + 1}{u}\, du}{2}$
    1. Rewrite the integrand:
      
      $\frac{u^{2} - 2 u + 1}{u} = u - 2 + \frac{1}{u}$
    2. 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 $\frac{du}{4}$ :
    
    $\int \frac{1}{16}\, du$
    1. 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:
    
    $\int \left(- 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\, dx = - 2 \int \tan{\left(x \right)} \sec^{2}{\left(x \right)}\, dx$
    1. Let $u = \sec^{2}{\left(x \right)}$ .
      
      Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{4}\, du$
      
      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 $- du$ :
    
    $\int \frac{1}{u}\, du$
    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: $- \log{\left(u \right)}$
    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 $\frac{du}{4}$ :
    
    $\int \frac{1}{16}\, du$
    1. 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:
    
    $\int \left(- 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)}\right)\, dx = - 2 \int \tan{\left(x \right)} \sec^{2}{\left(x \right)}\, dx$
    1. Let $u = \sec^{2}{\left(x \right)}$ .
      
      Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{4}\, du$
      
      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 $- du$ :
    
    $\int \frac{1}{u}\, du$
    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: $- \log{\left(u \right)}$
    Now substitute $u$ back in:
    
    $- \log{\left(\cos{\left(x \right)} \right)}$
  The result is: $- \log{\left(\cos{\left(x \right)} \right)} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                   
 |                       /   2   \                4   
 |    5               log\sec (x)/      2      sec (x)
 | tan (x)*1 dx = C + ------------ - sec (x) + -------
 |                         2                      4   
/

$${{4\,\sin ^2x-3}\over{4\,\sin ^4x-8\,\sin ^2x+4}}-{{\log \left( \sin ^2x-1\right)}\over{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                            2   
3                 -1 + 4*cos (1)
- - log(cos(1)) - --------------
4                        4      
                    4*cos (1)

$$-{{\log \left(1-\sin ^21\right)}\over{2}}-{{3}\over{4\,\sin ^41-8\, \sin ^21+4}}+{{\sin ^21}\over{\sin ^41-2\,\sin ^21+1}}+{{3}\over{4}}$$

                            2   
3                 -1 + 4*cos (1)
- - log(cos(1)) - --------------
4                        4      
                    4*cos (1)

$$- \frac{-1 + 4 \cos^{2}{\left(1 \right)}}{4 \cos^{4}{\left(1 \right)}} - \log{\left(\cos{\left(1 \right)} \right)} + \frac{3}{4}$$

Simplify

Numerical answer [src]

0.87365244751029

0.87365244751029

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of tan^5xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3