Integral of tan³xdx dx

The solution

You have entered [src]

  1             
  /             
 |              
 |     3        
 |  tan (x)*1 dx
 |              
/               
0

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

Integral(tan(x)^3*1, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                         
 |                       2         /   2   \
 |    3               sec (x)   log\sec (x)/
 | tan (x)*1 dx = C + ------- - ------------
 |                       2           2      
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

$${{\log \left(1-\sin ^21\right)}\over{2}}-{{1}\over{2\,\sin ^21-2}}- {{1}\over{2}}$$

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

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

Simplify

Numerical answer [src]

0.597132940021366

0.597132940021366

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of tan³xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3