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sin^5x/cos^4x
sinus of to the power of 5x divide by co sinus of e of to the power of 4x
sin5x/cos4x
sin⁵x/cos⁴x
sin^5x divide by cos^4x
sin^5x/cos^4xdx

Solving integrals/ sin^5x/cos^4x

Integral of sin^5x/cos^4x dx

The solution

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  1           
  /           
 |            
 |     5      
 |  sin (x)   
 |  ------- dx
 |     4      
 |  cos (x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\sin^{5}{\left(x \right)}}{\cos^{4}{\left(x \right)}}\, dx$$

Integral(sin(x)^5/(cos(x)^4), (x, 0, 1))

Detail solution

Rewrite the integrand:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$-{{6\,\cos ^2x-1}\over{3\,\cos ^3x}}-\cos x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                       2   
8            -1 + 6*cos (1)
- - cos(1) - --------------
3                   3      
               3*cos (1)

$$-\cos 1-{{2}\over{\cos 1}}+{{1}\over{3\,\cos ^31}}+{{8}\over{3}}$$

                       2   
8            -1 + 6*cos (1)
- - cos(1) - --------------
3                   3      
               3*cos (1)

$$- \frac{-1 + 6 \cos^{2}{\left(1 \right)}}{3 \cos^{3}{\left(1 \right)}} - \cos{\left(1 \right)} + \frac{8}{3}$$

Simplify

Numerical answer [src]

0.538067617028605

0.538067617028605

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin^5x/cos^4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3