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Integral of tan^6(x)×sec^4(x)
Identical expressions
tan^ six (x)×sec^ four (x)
tangent of to the power of 6(x)×sec to the power of 4(x)
tangent of to the power of six (x)×sec to the power of four (x)
tan6(x)×sec4(x)
tan6x×sec4x
tan⁶(x)×sec⁴(x)
tan^6x×sec^4x
tan^6(x)×sec^4(x)dx

Solving integrals/ tan^6(x)×sec^4(x)

Integral of tan^6(x)×sec^4(x) dx

The solution

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  1                   
  /                   
 |                    
 |     6       4      
 |  tan (x)*sec (x) dx
 |                    
/                     
0

$$\int\limits_{0}^{1} \tan^{6}{\left(x \right)} \sec^{4}{\left(x \right)}\, dx$$

Integral(tan(x)^6*sec(x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\tan^{6}{\left(x \right)} \sec^{4}{\left(x \right)} = \left(\tan^{2}{\left(x \right)} + 1\right) \tan^{6}{\left(x \right)} \sec^{2}{\left(x \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \tan{\left(x \right)}$ .
  
  Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$ :
  
  $\int \left(u^{8} + u^{6}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
    The result is: $\frac{u^{9}}{9} + \frac{u^{7}}{7}$
  Now substitute $u$ back in:
  
  $\frac{\tan^{9}{\left(x \right)}}{9} + \frac{\tan^{7}{\left(x \right)}}{7}$
Method #2
1. Rewrite the integrand:
  
  $\left(\tan^{2}{\left(x \right)} + 1\right) \tan^{6}{\left(x \right)} \sec^{2}{\left(x \right)} = \tan^{8}{\left(x \right)} \sec^{2}{\left(x \right)} + \tan^{6}{\left(x \right)} \sec^{2}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \tan{\left(x \right)}$ .
    
    Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$ :
    
    $\int u^{8}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
    Now substitute $u$ back in:
    
    $\frac{\tan^{9}{\left(x \right)}}{9}$
  1. Let $u = \tan{\left(x \right)}$ .
    
    Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$ :
    
    $\int u^{6}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
    Now substitute $u$ back in:
    
    $\frac{\tan^{7}{\left(x \right)}}{7}$
  The result is: $\frac{\tan^{9}{\left(x \right)}}{9} + \frac{\tan^{7}{\left(x \right)}}{7}$
Method #3
1. Rewrite the integrand:
  
  $\left(\tan^{2}{\left(x \right)} + 1\right) \tan^{6}{\left(x \right)} \sec^{2}{\left(x \right)} = \tan^{8}{\left(x \right)} \sec^{2}{\left(x \right)} + \tan^{6}{\left(x \right)} \sec^{2}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \tan{\left(x \right)}$ .
    
    Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$ :
    
    $\int u^{8}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
    Now substitute $u$ back in:
    
    $\frac{\tan^{9}{\left(x \right)}}{9}$
  1. Let $u = \tan{\left(x \right)}$ .
    
    Then let $du = \left(\tan^{2}{\left(x \right)} + 1\right) dx$ and substitute $du$ :
    
    $\int u^{6}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
    Now substitute $u$ back in:
    
    $\frac{\tan^{7}{\left(x \right)}}{7}$
  The result is: $\frac{\tan^{9}{\left(x \right)}}{9} + \frac{\tan^{7}{\left(x \right)}}{7}$
Add the constant of integration:

$\frac{\tan^{9}{\left(x \right)}}{9} + \frac{\tan^{7}{\left(x \right)}}{7}+ \mathrm{constant}$

The answer is:

$\frac{\tan^{9}{\left(x \right)}}{9} + \frac{\tan^{7}{\left(x \right)}}{7}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                          
 |                             7         9   
 |    6       4             tan (x)   tan (x)
 | tan (x)*sec (x) dx = C + ------- + -------
 |                             7         9   
/

$${{7\,\tan ^9x+9\,\tan ^7x}\over{63}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  19*sin(1)     2*sin(1)     sin(1)       sin(1)     5*sin(1) 
- ---------- - --------- - ---------- + --------- + ----------
        7      63*cos(1)         3           9            5   
  63*cos (1)               63*cos (1)   9*cos (1)   21*cos (1)

$${{7\,\tan ^91+9\,\tan ^71}\over{63}}$$

  19*sin(1)     2*sin(1)     sin(1)       sin(1)     5*sin(1) 
- ---------- - --------- - ---------- + --------- + ----------
        7      63*cos(1)         3           9            5   
  63*cos (1)               63*cos (1)   9*cos (1)   21*cos (1)

$$- \frac{19 \sin{\left(1 \right)}}{63 \cos^{7}{\left(1 \right)}} - \frac{\sin{\left(1 \right)}}{63 \cos^{3}{\left(1 \right)}} - \frac{2 \sin{\left(1 \right)}}{63 \cos{\left(1 \right)}} + \frac{5 \sin{\left(1 \right)}}{21 \cos^{5}{\left(1 \right)}} + \frac{\sin{\left(1 \right)}}{9 \cos^{9}{\left(1 \right)}}$$

Simplify

Numerical answer [src]

9.16414591049723

9.16414591049723

The graph

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

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How to use it?

Identical expressions

Integral of tan^6(x)×sec^4(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3