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Integral of d{x}:
Integral of (4x-x^2)dx
Integral of (2x+1)^3
Integral of x^5*dx
Integral of x/(x^3-3x+2)
Identical expressions
dx/sin^5xcos^3x
dx divide by sinus of to the power of 5x co sinus of e of cubed x
dx/sin5xcos3x
dx/sin⁵xcos³x
dx/sin to the power of 5xcos to the power of 3x
dx divide by sin^5xcos^3x

Solving integrals/ dx/sin^5xcos^3x

Integral of dx/sin^5xcos^3x dx

The solution

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  1           
  /           
 |            
 |     3      
 |  cos (x)   
 |  ------- dx
 |     5      
 |  sin (x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\cos^{3}{\left(x \right)}}{\sin^{5}{\left(x \right)}}\, dx$$

Integral(cos(x)^3/sin(x)^5, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$- \frac{\cos{\left(2 x \right)}}{4 \sin^{4}{\left(x \right)}}$
Add the constant of integration:

$- \frac{\cos{\left(2 x \right)}}{4 \sin^{4}{\left(x \right)}}+ \mathrm{constant}$

The answer is:

$- \frac{\cos{\left(2 x \right)}}{4 \sin^{4}{\left(x \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                      
 |                                       
 |    3                                  
 | cos (x)              1           1    
 | ------- dx = C + --------- - ---------
 |    5                  2           4   
 | sin (x)          2*sin (x)   4*sin (x)
 |                                       
/

$$\int \frac{\cos^{3}{\left(x \right)}}{\sin^{5}{\left(x \right)}}\, dx = C + \frac{1}{2 \sin^{2}{\left(x \right)}} - \frac{1}{4 \sin^{4}{\left(x \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

oo

Numerical answer [src]

7.26749061658134e+75

7.26749061658134e+75

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^5xcos^3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3