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Integral of d{x}:
Integral of 2x^4
Integral of x^3-3x^2
Integral of tan^3(x)sec^7(x)
Integral of (x^2-5x+9)/(x^2-5x+6)
Identical expressions
tan^ three (x)sec^ seven (x)
tangent of cubed (x)sec to the power of 7(x)
tangent of to the power of three (x)sec to the power of seven (x)
tan3(x)sec7(x)
tan3xsec7x
tan³(x)sec⁷(x)
tan to the power of 3(x)sec to the power of 7(x)
tan^3xsec^7x
tan^3(x)sec^7(x)dx

Solving integrals/ tan^3(x)sec^7(x)

Integral of tan^3(x)sec^7(x) dx

The solution

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  1                   
  /                   
 |                    
 |     3       7      
 |  tan (x)*sec (x) dx
 |                    
/                     
0

$$\int\limits_{0}^{1} \tan^{3}{\left(x \right)} \sec^{7}{\left(x \right)}\, dx$$

Integral(tan(x)^3*sec(x)^7, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\tan^{3}{\left(x \right)} \sec^{7}{\left(x \right)} = \left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{7}{\left(x \right)}$
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 $du$ :
  
  $\int \left(u^{8} - u^{6}\right)\, du$
  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 a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
    The result is: $\frac{u^{9}}{9} - \frac{u^{7}}{7}$
  Now substitute $u$ back in:
  
  $\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}$
Method #2
1. Rewrite the integrand:
  
  $\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{7}{\left(x \right)} = \tan{\left(x \right)} \sec^{9}{\left(x \right)} - \tan{\left(x \right)} \sec^{7}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sec^{9}{\left(x \right)}$ .
    
    Then let $du = 9 \tan{\left(x \right)} \sec^{9}{\left(x \right)} dx$ and substitute $\frac{du}{9}$ :
    
    $\int \frac{1}{81}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{9}\, du = \frac{\int 1\, du}{9}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{9}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \tan{\left(x \right)} \sec^{7}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{7}{\left(x \right)}\, dx$
    1. Let $u = \sec^{7}{\left(x \right)}$ .
      
      Then let $du = 7 \tan{\left(x \right)} \sec^{7}{\left(x \right)} dx$ and substitute $\frac{du}{7}$ :
      
      $\int \frac{1}{49}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{7}\, du = \frac{\int 1\, du}{7}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{7}{\left(x \right)}}{7}$
    So, the result is: $- \frac{\sec^{7}{\left(x \right)}}{7}$
  The result is: $\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}$
Method #3
1. Rewrite the integrand:
  
  $\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{7}{\left(x \right)} = \tan{\left(x \right)} \sec^{9}{\left(x \right)} - \tan{\left(x \right)} \sec^{7}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sec^{9}{\left(x \right)}$ .
    
    Then let $du = 9 \tan{\left(x \right)} \sec^{9}{\left(x \right)} dx$ and substitute $\frac{du}{9}$ :
    
    $\int \frac{1}{81}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{9}\, du = \frac{\int 1\, du}{9}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{9}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \tan{\left(x \right)} \sec^{7}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{7}{\left(x \right)}\, dx$
    1. Let $u = \sec^{7}{\left(x \right)}$ .
      
      Then let $du = 7 \tan{\left(x \right)} \sec^{7}{\left(x \right)} dx$ and substitute $\frac{du}{7}$ :
      
      $\int \frac{1}{49}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{7}\, du = \frac{\int 1\, du}{7}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{7}{\left(x \right)}}{7}$
    So, the result is: $- \frac{\sec^{7}{\left(x \right)}}{7}$
  The result is: $\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                          
 |                             7         9   
 |    3       7             sec (x)   sec (x)
 | tan (x)*sec (x) dx = C - ------- + -------
 |                             7         9   
/

$$-{{9\,\cos ^2x-7}\over{63\,\cos ^9x}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

              2   
2    7 - 9*cos (1)
-- + -------------
63           9    
       63*cos (1)

$$-{{1}\over{7\,\cos ^71}}+{{1}\over{9\,\cos ^91}}+{{2}\over{63}}$$

              2   
2    7 - 9*cos (1)
-- + -------------
63           9    
       63*cos (1)

$$\frac{2}{63} + \frac{7 - 9 \cos^{2}{\left(1 \right)}}{63 \cos^{9}{\left(1 \right)}}$$

Упростить

Numerical answer [src]

17.7195471832138

17.7195471832138

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^3(x)sec^7(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3