Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x/(1-x^2)^1/2
Integral of x^2*e^(4*x)
Integral of x*(-2)/(1+x^2)
Integral of x^2√(1-x^2)
Identical expressions
dx/sinxcos^3x
dx divide by sinus of x co sinus of e of cubed x
dx/sinxcos3x
dx/sinxcos³x
dx/sinxcos to the power of 3x
dx divide by sinxcos^3x

Integral of dx/sinxcos^3x dx

The solution

You have entered [src]

 pi           
 --           
 3            
  /           
 |            
 |     3      
 |  cos (x)   
 |  ------- dx
 |   sin(x)   
 |            
/             
pi            
--            
4

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

Integral(cos(x)^3/sin(x), (x, pi/4, pi/3))

Detail solution

Rewrite the integrand:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                       
 |                                        
 |    3                /   2   \      2   
 | cos (x)          log\sin (x)/   sin (x)
 | ------- dx = C + ------------ - -------
 |  sin(x)               2            2   
 |                                        
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

         /  ___\      /  ___\
  1      |\/ 2 |      |\/ 3 |
- - - log|-----| + log|-----|
  8      \  2  /      \  2  /

$$\log{\left(\frac{\sqrt{3}}{2} \right)} - \frac{1}{8} - \log{\left(\frac{\sqrt{2}}{2} \right)}$$

         /  ___\      /  ___\
  1      |\/ 2 |      |\/ 3 |
- - - log|-----| + log|-----|
  8      \  2  /      \  2  /

$$\log{\left(\frac{\sqrt{3}}{2} \right)} - \frac{1}{8} - \log{\left(\frac{\sqrt{2}}{2} \right)}$$

-1/8 - log(sqrt(2)/2) + log(sqrt(3)/2)

Numerical answer [src]

0.0777325540540822

0.0777325540540822

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

Other calculators

How to use it?

Identical expressions

Integral of dx/sinxcos^3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3