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Integral of d{x}:
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Integral of (1-cosx)/(1+cosx)
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Integral of xcos(5x)
Identical expressions
dx/(sin^6x)
dx divide by ( sinus of to the power of 6x)
dx/(sin6x)
dx/sin6x
dx/(sin⁶x)
dx/sin^6x
dx divide by (sin^6x)

Solving integrals/ dx/(sin^6x)

Integral of dx/(sin^6x) dx

The solution

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  1             
  /             
 |              
 |       1      
 |  1*------- dx
 |       6      
 |    sin (x)   
 |              
/               
0

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

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

Detail solution

Rewrite the integrand:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                               
 |                                  3         5   
 |      1                      2*cot (x)   cot (x)
 | 1*------- dx = C - cot(x) - --------- - -------
 |      6                          3          5   
 |   sin (x)                                      
 |                                                
/

$$-{{15\,\tan ^4x+10\,\tan ^2x+3}\over{15\,\tan ^5x}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$${\it \%a}$$

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.

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How to use it?

Identical expressions

Integral of dx/(sin^6x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2