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Identical expressions
tg^3xdx/cos^2x
tg cubed xdx divide by co sinus of e of squared x
tg3xdx/cos2x
tg³xdx/cos²x
tg to the power of 3xdx/cos to the power of 2x
tg^3xdx divide by cos^2x

Solving integrals/ tg^3xdx/cos^2x

Integral of tg^3xdx/cos^2x dx

The solution

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  1                     
  /                     
 |                      
 |     3         1      
 |  tan (x)*1*------- dx
 |               2      
 |            cos (x)   
 |                      
/                       
0

$$\int\limits_{0}^{1} \tan^{3}{\left(x \right)} 1 \cdot \frac{1}{\cos^{2}{\left(x \right)}}\, dx$$

Integral(tan(x)^3*1/cos(x)^2, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$\frac{\left(\sec^{2}{\left(x \right)} - 2\right) \sec^{2}{\left(x \right)}}{4}$
Add the constant of integration:

$\frac{\left(\sec^{2}{\left(x \right)} - 2\right) \sec^{2}{\left(x \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{\left(\sec^{2}{\left(x \right)} - 2\right) \sec^{2}{\left(x \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                               2         4   
 |    3         1             sec (x)   sec (x)
 | tan (x)*1*------- dx = C - ------- + -------
 |              2                2         4   
 |           cos (x)                           
 |                                             
/

$${{2\,\sin ^2x-1}\over{4\,\sin ^4x-8\,\sin ^2x+4}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

             2   
1   1 - 2*cos (1)
- + -------------
4          4     
      4*cos (1)

$$-{{1}\over{4\,\sin ^41-8\,\sin ^21+4}}+{{\sin ^21}\over{2\,\sin ^41 -4\,\sin ^21+2}}+{{1}\over{4}}$$

             2   
1   1 - 2*cos (1)
- + -------------
4          4     
      4*cos (1)

$$\frac{1}{4} + \frac{1 - 2 \cos^{2}{\left(1 \right)}}{4 \cos^{4}{\left(1 \right)}}$$

Упростить

Numerical answer [src]

1.47078538753166

1.47078538753166

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 tg^3xdx/cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3