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Solving integrals/ (tg^3x-sin)/cos^2x

Integral of (tg^3x-sin)/cos^2x dx

The solution

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

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

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

Detail solution

Rewrite the integrand:

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

$\frac{-4 - \frac{2}{\cos{\left(x \right)}} + \frac{1}{\cos^{3}{\left(x \right)}}}{4 \cos{\left(x \right)}}$
Add the constant of integration:

$\frac{-4 - \frac{2}{\cos{\left(x \right)}} + \frac{1}{\cos^{3}{\left(x \right)}}}{4 \cos{\left(x \right)}}+ \mathrm{constant}$

The answer is:

$\frac{-4 - \frac{2}{\cos{\left(x \right)}} + \frac{1}{\cos^{3}{\left(x \right)}}}{4 \cos{\left(x \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\int \frac{- \sin{\left(x \right)} + \tan^{3}{\left(x \right)}}{\cos^{2}{\left(x \right)}}\, dx = C + \frac{\sec^{4}{\left(x \right)}}{4} - \frac{\sec^{2}{\left(x \right)}}{2} - \frac{1}{\cos{\left(x \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                2         4   
5     1      sec (1)   sec (1)
- - ------ - ------- + -------
4   cos(1)      2         4

$$- \frac{1}{\cos{\left(1 \right)}} - \frac{\sec^{2}{\left(1 \right)}}{2} + \frac{5}{4} + \frac{\sec^{4}{\left(1 \right)}}{4}$$

                2         4   
5     1      sec (1)   sec (1)
- - ------ - ------- + -------
4   cos(1)      2         4

$$- \frac{1}{\cos{\left(1 \right)}} - \frac{\sec^{2}{\left(1 \right)}}{2} + \frac{5}{4} + \frac{\sec^{4}{\left(1 \right)}}{4}$$

Упростить

Numerical answer [src]

0.61996966985073

0.61996966985073

The graph

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

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How to use it?

Identical expressions

Similar expressions

Integral of (tg^3x-sin)/cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3