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Identical expressions
cos^5x/sin^2x
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cos^5x/sin^2xdx

Solving integrals/ cos^5x/sin^2x

Integral of cos^5x/sin^2x dx

The solution

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  1           
  /           
 |            
 |     5      
 |  cos (x)   
 |  ------- dx
 |     2      
 |  sin (x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\cos^{5}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx$$

Integral(cos(x)^5/(sin(x)^2), (x, 0, 1))

Detail solution

Rewrite the integrand:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                            
 |                                             
 |    5                                    3   
 | cos (x)            1                 sin (x)
 | ------- dx = C - ------ - 2*sin(x) + -------
 |    2             sin(x)                 3   
 | sin (x)                                     
 |                                             
/

$${{\sin ^3x-6\,\sin x}\over{3}}-{{1}\over{\sin x}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$${\it \%a}$$

oo

$$\infty$$

Simplify

Numerical answer [src]

1.3793236779486e+19

1.3793236779486e+19

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of cos^5x/sin^2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3