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cos^3xsin^2xdx
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Integral of cos^3xsin^2xdx dx

The solution

You have entered [src]

  p                     
  -                     
  2                     
  /                     
 |                      
 |     3       2        
 |  cos (x)*sin (x)*1 dx
 |                      
/                       
0

$$\int\limits_{0}^{\frac{p}{2}} \cos^{3}{\left(x \right)} \sin^{2}{\left(x \right)} 1\, dx$$

Integral(cos(x)^3*sin(x)^2*1, (x, 0, p/2))

Detail solution

Rewrite the integrand:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$-{{3\,\sin ^5x-5\,\sin ^3x}\over{15}}$$

Simplify

The answer [src]

     5/p\      3/p\
  sin |-|   sin |-|
      \2/       \2/
- ------- + -------
     5         3

$$- \frac{\sin^{5}{\left(\frac{p}{2} \right)}}{5} + \frac{\sin^{3}{\left(\frac{p}{2} \right)}}{3}$$

     5/p\      3/p\
  sin |-|   sin |-|
      \2/       \2/
- ------- + -------
     5         3

$$- \frac{\sin^{5}{\left(\frac{p}{2} \right)}}{5} + \frac{\sin^{3}{\left(\frac{p}{2} \right)}}{3}$$

Упростить

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

Other calculators

How to use it?

Identical expressions

Integral of cos^3xsin^2xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3