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Identical expressions
tan^3xsec^4xdx
tangent of cubed xsec to the power of 4xdx
tan3xsec4xdx
tan³xsec⁴xdx
tan to the power of 3xsec to the power of 4xdx

Integral of tan^3xsec^4xdx dx

The solution

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  1                     
  /                     
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 |     3       4        
 |  tan (x)*sec (x)*1 dx
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/                       
0

$$\int\limits_{0}^{1} \tan^{3}{\left(x \right)} \sec^{4}{\left(x \right)} 1\, dx$$

Integral(tan(x)^3*sec(x)^4*1, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$\frac{\sec^{6}{\left(x \right)}}{6} - \frac{\sec^{4}{\left(x \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{\sec^{6}{\left(x \right)}}{6} - \frac{\sec^{4}{\left(x \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                               4         6   
 |    3       4               sec (x)   sec (x)
 | tan (x)*sec (x)*1 dx = C - ------- + -------
 |                               4         6   
/

$$-{{3\,\sin ^2x-1}\over{12\,\sin ^6x-36\,\sin ^4x+36\,\sin ^2x-12}}$$

Simplify

The answer [src]

              2   
1    2 - 3*cos (1)
-- + -------------
12           6    
       12*cos (1)

$${{1}\over{12\,\sin ^61-36\,\sin ^41+36\,\sin ^21-12}}-{{\sin ^21 }\over{4\,\sin ^61-12\,\sin ^41+12\,\sin ^21-4}}+{{1}\over{12}}$$

              2   
1    2 - 3*cos (1)
-- + -------------
12           6    
       12*cos (1)

$$\frac{1}{12} + \frac{2 - 3 \cos^{2}{\left(1 \right)}}{12 \cos^{6}{\left(1 \right)}}$$

Упростить

Numerical answer [src]

3.84906381342323

3.84906381342323

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

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Identical expressions

Integral of tan^3xsec^4xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3