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Integral of d{x}:
Integral of xe
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Integral of e^-t
Integral of e^(x*y)
Identical expressions
(cos^3x/sin^6x)
( co sinus of e of cubed x divide by sinus of to the power of 6x)
(cos3x/sin6x)
cos3x/sin6x
(cos³x/sin⁶x)
(cos to the power of 3x/sin to the power of 6x)
cos^3x/sin^6x
(cos^3x divide by sin^6x)
(cos^3x/sin^6x)dx

Solving integrals/ (cos^3x/sin^6x)

Integral of (cos^3x/sin^6x) dx

The solution

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  1           
  /           
 |            
 |     3      
 |  cos (x)   
 |  ------- dx
 |     6      
 |  sin (x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\cos^{3}{\left(x \right)}}{\sin^{6}{\left(x \right)}}\, dx$$

Integral(cos(x)^3/(sin(x)^6), (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$\frac{5 \sin^{2}{\left(x \right)} - 3}{15 \sin^{5}{\left(x \right)}}$
Add the constant of integration:

$\frac{5 \sin^{2}{\left(x \right)} - 3}{15 \sin^{5}{\left(x \right)}}+ \mathrm{constant}$

The answer is:

$\frac{5 \sin^{2}{\left(x \right)} - 3}{15 \sin^{5}{\left(x \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                      
 |                                       
 |    3                                  
 | cos (x)              1           1    
 | ------- dx = C - --------- + ---------
 |    6                  5           3   
 | sin (x)          5*sin (x)   3*sin (x)
 |                                       
/

$$\int \frac{\cos^{3}{\left(x \right)}}{\sin^{6}{\left(x \right)}}\, dx = C + \frac{1}{3 \sin^{3}{\left(x \right)}} - \frac{1}{5 \sin^{5}{\left(x \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

Упростить

Numerical answer [src]

7.0110751903966e+94

7.0110751903966e+94

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of (cos^3x/sin^6x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3