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sin^ two *(2x)cosx
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sin^2*(2x)cosxdx

Integral of sin^2*(2x)cosx dx

The solution

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

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

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

Detail solution

Rewrite the integrand:

$\sin^{2}{\left(2 x \right)} \cos{\left(x \right)} = 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}$
The integral of a constant times a function is the constant times the integral of the function:

$\int 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = 4 \int \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$
1. Rewrite the integrand:
  
  $\sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \sin^{2}{\left(x \right)} \cos{\left(x \right)}$
2. 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:
      
      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: $- \frac{u^{5}}{5}$
      
      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:
      
      $\int \left(- \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      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 $du$ :
      
      $\int u^{2}\, du$
      
      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:
      
      $\int \left(- \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      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 $du$ :
      
      $\int u^{2}\, du$
      
      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: $- \frac{\sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}$
So, the result is: $- \frac{4 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3}$
Now simplify:

$\frac{4 \left(5 - 3 \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}}{15}$
Add the constant of integration:

$\frac{4 \left(5 - 3 \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}}{15}+ \mathrm{constant}$

The answer is:

$\frac{4 \left(5 - 3 \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}}{15}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\int \sin^{2}{\left(2 x \right)} \cos{\left(x \right)}\, dx = C - \frac{4 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     2                  2                                   
7*sin (2)*sin(1)   8*cos (2)*sin(1)   4*cos(1)*cos(2)*sin(2)
---------------- + ---------------- - ----------------------
       15                 15                    15

$$- \frac{4 \sin{\left(2 \right)} \cos{\left(1 \right)} \cos{\left(2 \right)}}{15} + \frac{8 \sin{\left(1 \right)} \cos^{2}{\left(2 \right)}}{15} + \frac{7 \sin{\left(1 \right)} \sin^{2}{\left(2 \right)}}{15}$$

     2                  2                                   
7*sin (2)*sin(1)   8*cos (2)*sin(1)   4*cos(1)*cos(2)*sin(2)
---------------- + ---------------- - ----------------------
       15                 15                    15

7*sin(2)^2*sin(1)/15 + 8*cos(2)^2*sin(1)/15 - 4*cos(1)*cos(2)*sin(2)/15

Numerical answer [src]

0.45692170546545

0.45692170546545

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

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How to use it?

Identical expressions

Integral of sin^2*(2x)cosx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3