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(sin³x)/(cos⁴x)
( sinus of ³x) divide by ( co sinus of e of ⁴x)
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(sin³x)/(cos⁴x)dx

Integral of (sin³x)/(cos⁴x) dx

The solution

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  1           
  /           
 |            
 |     3      
 |  sin (x)   
 |  ------- dx
 |     4      
 |  cos (x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\sin^{3}{\left(x \right)}}{\cos^{4}{\left(x \right)}}\, dx$$

Integral(sin(x)^3/cos(x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                   
 |                                    
 |    3                               
 | sin (x)            1          1    
 | ------- dx = C - ------ + ---------
 |    4             cos(x)        3   
 | cos (x)                   3*cos (x)
 |                                    
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

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

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

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

2/3 - (-1 + 3*cos(1)^2)/(3*cos(1)^3)

Numerical answer [src]

0.92918564057767

0.92918564057767

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

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How to use it?

Identical expressions

Integral of (sin³x)/(cos⁴x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3