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Identical expressions
sin^6xcos^3x
sinus of to the power of 6x co sinus of e of cubed x
sin6xcos3x
sin⁶xcos³x
sin to the power of 6xcos to the power of 3x
sin^6xcos^3xdx

Solving integrals/ sin^6xcos^3x

Integral of sin^6xcos^3x dx

The solution

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  1                   
  /                   
 |                    
 |     6       3      
 |  sin (x)*cos (x) dx
 |                    
/                     
0

$$\int\limits_{0}^{1} \sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$

Integral(sin(x)^6*cos(x)^3, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$- \frac{\sin^{9}{\left(x \right)}}{9} + \frac{\sin^{7}{\left(x \right)}}{7}+ \mathrm{constant}$

The answer is:

$- \frac{\sin^{9}{\left(x \right)}}{9} + \frac{\sin^{7}{\left(x \right)}}{7}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                          
 |                             9         7   
 |    6       3             sin (x)   sin (x)
 | sin (x)*cos (x) dx = C - ------- + -------
 |                             9         7   
/

$$\int \sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = C - \frac{\sin^{9}{\left(x \right)}}{9} + \frac{\sin^{7}{\left(x \right)}}{7}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     9         7   
  sin (1)   sin (1)
- ------- + -------
     9         7

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

     9         7   
  sin (1)   sin (1)
- ------- + -------
     9         7

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

-sin(1)^9/9 + sin(1)^7/7

Numerical answer [src]

0.019172971209824

0.019172971209824

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^6xcos^3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3