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tan^ three (x)*sec^ three (x)
tangent of cubed (x) multiply by sec cubed (x)
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tan^3(x)sec^3(x)
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tan^3(x)*sec^3(x)dx

Solving integrals/ tan^3(x)*sec^3(x)

Integral of tan^3(x)*sec^3(x) dx

The solution

You have entered [src]

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

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

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

Detail solution

Rewrite the integrand:

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

$\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}+ \mathrm{constant}$

The answer is:

$\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                          
 |                             3         5   
 |    3       3             sec (x)   sec (x)
 | tan (x)*sec (x) dx = C - ------- + -------
 |                             3         5   
/

$$\int \tan^{3}{\left(x \right)} \sec^{3}{\left(x \right)}\, dx = C + \frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

               2   
2    -3 + 5*cos (1)
-- - --------------
15           5     
       15*cos (1)

$$\frac{2}{15} - \frac{-3 + 5 \cos^{2}{\left(1 \right)}}{15 \cos^{5}{\left(1 \right)}}$$

               2   
2    -3 + 5*cos (1)
-- - --------------
15           5     
       15*cos (1)

$$\frac{2}{15} - \frac{-3 + 5 \cos^{2}{\left(1 \right)}}{15 \cos^{5}{\left(1 \right)}}$$

2/15 - (-3 + 5*cos(1)^2)/(15*cos(1)^5)

Numerical answer [src]

2.36355929817875

2.36355929817875

The graph

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

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Identical expressions

Integral of tan^3(x)*sec^3(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3