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Integral of d{x}:
Integral of x/(x-1)^2
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Identical expressions
tgx^7sec^4x
tgx to the power of 7sec to the power of 4x
tgx7sec4x
tgx⁷sec⁴x
tgx^7sec^4xdx

Solving integrals/ tgx^7sec^4x

Integral of tgx^7sec^4x dx

The solution

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  1                   
  /                   
 |                    
 |     7       4      
 |  tan (x)*sec (x) dx
 |                    
/                     
0

$$\int\limits_{0}^{1} \tan^{7}{\left(x \right)} \sec^{4}{\left(x \right)}\, dx$$

Integral(tan(x)^7*sec(x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\tan^{7}{\left(x \right)} \sec^{4}{\left(x \right)} = \left(\sec^{2}{\left(x \right)} - 1\right)^{3} \tan{\left(x \right)} \sec^{4}{\left(x \right)}$
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 $du$ :
  
  $\int \left(\frac{u^{4}}{2} - \frac{3 u^{3}}{2} + \frac{3 u^{2}}{2} - \frac{u}{2}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $\frac{u^{5}}{10}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 u^{3}}{2}\right)\, du = - \frac{3 \int u^{3}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $- \frac{3 u^{4}}{8}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 u^{2}}{2}\, du = \frac{3 \int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u}{2}\right)\, du = - \frac{\int u\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{4}$
    The result is: $\frac{u^{5}}{10} - \frac{3 u^{4}}{8} + \frac{u^{3}}{2} - \frac{u^{2}}{4}$
  Now substitute $u$ back in:
  
  $\frac{\sec^{10}{\left(x \right)}}{10} - \frac{3 \sec^{8}{\left(x \right)}}{8} + \frac{\sec^{6}{\left(x \right)}}{2} - \frac{\sec^{4}{\left(x \right)}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\left(\sec^{2}{\left(x \right)} - 1\right)^{3} \tan{\left(x \right)} \sec^{4}{\left(x \right)} = \tan{\left(x \right)} \sec^{10}{\left(x \right)} - 3 \tan{\left(x \right)} \sec^{8}{\left(x \right)} + 3 \tan{\left(x \right)} \sec^{6}{\left(x \right)} - \tan{\left(x \right)} \sec^{4}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sec^{10}{\left(x \right)}$ .
    
    Then let $du = 10 \tan{\left(x \right)} \sec^{10}{\left(x \right)} dx$ and substitute $\frac{du}{10}$ :
    
    $\int \frac{1}{100}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{10}\, du = \frac{\int 1\, du}{10}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{10}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{10}{\left(x \right)}}{10}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 \tan{\left(x \right)} \sec^{8}{\left(x \right)}\right)\, dx = - 3 \int \tan{\left(x \right)} \sec^{8}{\left(x \right)}\, dx$
    1. Let $u = \sec^{8}{\left(x \right)}$ .
      
      Then let $du = 8 \tan{\left(x \right)} \sec^{8}{\left(x \right)} dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{1}{64}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8}\, du = \frac{\int 1\, du}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{8}{\left(x \right)}}{8}$
    So, the result is: $- \frac{3 \sec^{8}{\left(x \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 \tan{\left(x \right)} \sec^{6}{\left(x \right)}\, dx = 3 \int \tan{\left(x \right)} \sec^{6}{\left(x \right)}\, dx$
    1. Let $u = \sec^{6}{\left(x \right)}$ .
      
      Then let $du = 6 \tan{\left(x \right)} \sec^{6}{\left(x \right)} dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{1}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{6}\, du = \frac{\int 1\, du}{6}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{6}{\left(x \right)}}{6}$
    So, the result is: $\frac{\sec^{6}{\left(x \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \tan{\left(x \right)} \sec^{4}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{4}{\left(x \right)}\, dx$
    1. Let $u = \sec^{4}{\left(x \right)}$ .
      
      Then let $du = 4 \tan{\left(x \right)} \sec^{4}{\left(x \right)} dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{16}\, du$
      
      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:
      
      $\frac{\sec^{4}{\left(x \right)}}{4}$
    So, the result is: $- \frac{\sec^{4}{\left(x \right)}}{4}$
  The result is: $\frac{\sec^{10}{\left(x \right)}}{10} - \frac{3 \sec^{8}{\left(x \right)}}{8} + \frac{\sec^{6}{\left(x \right)}}{2} - \frac{\sec^{4}{\left(x \right)}}{4}$
Method #3
1. Rewrite the integrand:
  
  $\left(\sec^{2}{\left(x \right)} - 1\right)^{3} \tan{\left(x \right)} \sec^{4}{\left(x \right)} = \tan{\left(x \right)} \sec^{10}{\left(x \right)} - 3 \tan{\left(x \right)} \sec^{8}{\left(x \right)} + 3 \tan{\left(x \right)} \sec^{6}{\left(x \right)} - \tan{\left(x \right)} \sec^{4}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sec^{10}{\left(x \right)}$ .
    
    Then let $du = 10 \tan{\left(x \right)} \sec^{10}{\left(x \right)} dx$ and substitute $\frac{du}{10}$ :
    
    $\int \frac{1}{100}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{10}\, du = \frac{\int 1\, du}{10}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{10}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{10}{\left(x \right)}}{10}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 \tan{\left(x \right)} \sec^{8}{\left(x \right)}\right)\, dx = - 3 \int \tan{\left(x \right)} \sec^{8}{\left(x \right)}\, dx$
    1. Let $u = \sec^{8}{\left(x \right)}$ .
      
      Then let $du = 8 \tan{\left(x \right)} \sec^{8}{\left(x \right)} dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{1}{64}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{8}\, du = \frac{\int 1\, du}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{8}{\left(x \right)}}{8}$
    So, the result is: $- \frac{3 \sec^{8}{\left(x \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 \tan{\left(x \right)} \sec^{6}{\left(x \right)}\, dx = 3 \int \tan{\left(x \right)} \sec^{6}{\left(x \right)}\, dx$
    1. Let $u = \sec^{6}{\left(x \right)}$ .
      
      Then let $du = 6 \tan{\left(x \right)} \sec^{6}{\left(x \right)} dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{1}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{6}\, du = \frac{\int 1\, du}{6}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{6}{\left(x \right)}}{6}$
    So, the result is: $\frac{\sec^{6}{\left(x \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \tan{\left(x \right)} \sec^{4}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{4}{\left(x \right)}\, dx$
    1. Let $u = \sec^{4}{\left(x \right)}$ .
      
      Then let $du = 4 \tan{\left(x \right)} \sec^{4}{\left(x \right)} dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{1}{16}\, du$
      
      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:
      
      $\frac{\sec^{4}{\left(x \right)}}{4}$
    So, the result is: $- \frac{\sec^{4}{\left(x \right)}}{4}$
  The result is: $\frac{\sec^{10}{\left(x \right)}}{10} - \frac{3 \sec^{8}{\left(x \right)}}{8} + \frac{\sec^{6}{\left(x \right)}}{2} - \frac{\sec^{4}{\left(x \right)}}{4}$
Now simplify:

$\frac{\left(4 \sec^{6}{\left(x \right)} - 15 \sec^{4}{\left(x \right)} + 20 \sec^{2}{\left(x \right)} - 10\right) \sec^{4}{\left(x \right)}}{40}$
Add the constant of integration:

$\frac{\left(4 \sec^{6}{\left(x \right)} - 15 \sec^{4}{\left(x \right)} + 20 \sec^{2}{\left(x \right)} - 10\right) \sec^{4}{\left(x \right)}}{40}+ \mathrm{constant}$

The answer is:

$\frac{\left(4 \sec^{6}{\left(x \right)} - 15 \sec^{4}{\left(x \right)} + 20 \sec^{2}{\left(x \right)} - 10\right) \sec^{4}{\left(x \right)}}{40}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                 
 |                             6           8         4         10   
 |    7       4             sec (x)   3*sec (x)   sec (x)   sec  (x)
 | tan (x)*sec (x) dx = C + ------- - --------- - ------- + --------
 |                             2          8          4         10   
/

$$\int \tan^{7}{\left(x \right)} \sec^{4}{\left(x \right)}\, dx = C + \frac{\sec^{10}{\left(x \right)}}{10} - \frac{3 \sec^{8}{\left(x \right)}}{8} + \frac{\sec^{6}{\left(x \right)}}{2} - \frac{\sec^{4}{\left(x \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

               2            6            4   
1    4 - 15*cos (1) - 10*cos (1) + 20*cos (1)
-- + ----------------------------------------
40                       10                  
                   40*cos  (1)

$$\frac{1}{40} + \frac{- 15 \cos^{2}{\left(1 \right)} - 10 \cos^{6}{\left(1 \right)} + 20 \cos^{4}{\left(1 \right)} + 4}{40 \cos^{10}{\left(1 \right)}}$$

               2            6            4   
1    4 - 15*cos (1) - 10*cos (1) + 20*cos (1)
-- + ----------------------------------------
40                       10                  
                   40*cos  (1)

$$\frac{1}{40} + \frac{- 15 \cos^{2}{\left(1 \right)} - 10 \cos^{6}{\left(1 \right)} + 20 \cos^{4}{\left(1 \right)} + 4}{40 \cos^{10}{\left(1 \right)}}$$

Упростить

Numerical answer [src]

12.7214684874328

12.7214684874328

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 tgx^7sec^4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3