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Identical expressions
sin^4x*cos^5x
sinus of to the power of 4x multiply by co sinus of e of to the power of 5x
sin4x*cos5x
sin⁴x*cos⁵x
sin^4xcos^5x
sin4xcos5x
sin^4x*cos^5xdx

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

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

The solution

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

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

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

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{4}{\left(x \right)} \cos{\left(x \right)}$
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 $du$ :
  
  $\int \left(u^{8} - 2 u^{6} + u^{4}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u^{6}\right)\, du = - 2 \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{2 u^{7}}{7}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
    The result is: $\frac{u^{9}}{9} - \frac{2 u^{7}}{7} + \frac{u^{5}}{5}$
  Now substitute $u$ back in:
  
  $\frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{4}{\left(x \right)} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + \sin^{4}{\left(x \right)} \cos{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{8}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(x \right)}}{7}$
    So, the result is: $- \frac{2 \sin^{7}{\left(x \right)}}{7}$
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{4}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{5}{\left(x \right)}}{5}$
  The result is: $\frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{4}{\left(x \right)} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + \sin^{4}{\left(x \right)} \cos{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{8}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(x \right)}}{7}$
    So, the result is: $- \frac{2 \sin^{7}{\left(x \right)}}{7}$
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{4}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{5}{\left(x \right)}}{5}$
  The result is: $\frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}$
Now simplify:

$\frac{\left(35 \sin^{4}{\left(x \right)} - 90 \sin^{2}{\left(x \right)} + 63\right) \sin^{5}{\left(x \right)}}{315}$
Add the constant of integration:

$\frac{\left(35 \sin^{4}{\left(x \right)} - 90 \sin^{2}{\left(x \right)} + 63\right) \sin^{5}{\left(x \right)}}{315}+ \mathrm{constant}$

The answer is:

$\frac{\left(35 \sin^{4}{\left(x \right)} - 90 \sin^{2}{\left(x \right)} + 63\right) \sin^{5}{\left(x \right)}}{315}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                      
 |                               7         5         9   
 |    4       5             2*sin (x)   sin (x)   sin (x)
 | sin (x)*cos (x) dx = C - --------- + ------- + -------
 |                              7          5         9   
/

$$\int \sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx = C + \frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       7         5         9   
  2*sin (1)   sin (1)   sin (1)
- --------- + ------- + -------
      7          5         9

$$- \frac{2 \sin^{7}{\left(1 \right)}}{7} + \frac{\sin^{9}{\left(1 \right)}}{9} + \frac{\sin^{5}{\left(1 \right)}}{5}$$

       7         5         9   
  2*sin (1)   sin (1)   sin (1)
- --------- + ------- + -------
      7          5         9

$$- \frac{2 \sin^{7}{\left(1 \right)}}{7} + \frac{\sin^{9}{\left(1 \right)}}{9} + \frac{\sin^{5}{\left(1 \right)}}{5}$$

-2*sin(1)^7/7 + sin(1)^5/5 + sin(1)^9/9

Numerical answer [src]

0.0225291073790017

0.0225291073790017

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 sin^4x*cos^5x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3