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cos^3x/sin^4x
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Similar expressions
cos^3(x)/sin^4(x)

Solving integrals/ cos^3x/sin^4x

Integral of cos^3x/sin^4x dx

The solution

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

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

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

Detail solution

Rewrite the integrand:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$${{3\,\sin ^2x-1}\over{3\,\sin ^3x}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$${\it \%a}$$

oo

$$\infty$$

Simplify

Numerical answer [src]

7.81431122445857e+56

7.81431122445857e+56

The graph

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

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Identical expressions

Similar expressions

Integral of cos^3x/sin^4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3