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Identical expressions
sin^5x/cos^2x
sinus of to the power of 5x divide by co sinus of e of squared x
sin5x/cos2x
sin⁵x/cos²x
sin to the power of 5x/cos to the power of 2x
sin^5x divide by cos^2x
sin^5x/cos^2xdx

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

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

The solution

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

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

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

Detail solution

Rewrite the integrand:

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

$\frac{\left(\sin^{2}{\left(x \right)} + 5\right) \cos^{2}{\left(x \right)} + 3}{3 \cos{\left(x \right)}}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

$$-{{\cos ^31-6\,\cos 1}\over{3}}+{{1}\over{\cos 1}}-{{8}\over{3}}$$

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

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

Simplify

Numerical answer [src]

0.212177461000207

0.212177461000207

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^2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3