Integral of sin3xcos2xdx dx

The solution

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

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

Detail solution

Rewrite the integrand:

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

$- \frac{8 \cos^{5}{\left(x \right)}}{5} + 2 \cos^{3}{\left(x \right)} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer is:

$- \frac{8 \cos^{5}{\left(x \right)}}{5} + 2 \cos^{3}{\left(x \right)} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     5   
 |                                          3      8*cos (x)
 | sin(3*x)*cos(2*x) dx = C - cos(x) + 2*cos (x) - ---------
 |                                                     5    
/

$$\int \sin{\left(3 x \right)} \cos{\left(2 x \right)}\, dx = C - \frac{8 \cos^{5}{\left(x \right)}}{5} + 2 \cos^{3}{\left(x \right)} - \cos{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

3   3*cos(2)*cos(3)   2*sin(2)*sin(3)
- - --------------- - ---------------
5          5                 5

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

3   3*cos(2)*cos(3)   2*sin(2)*sin(3)
- - --------------- - ---------------
5          5                 5

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

3/5 - 3*cos(2)*cos(3)/5 - 2*sin(2)*sin(3)/5

Numerical answer [src]

0.301482628519607

0.301482628519607

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin3xcos2xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3