Integral of sin2x*sinx*cos3x dx

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 |  sin(2*x)*sin(x)*cos(3*x) dx
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0

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} \cos{\left(3 x \right)}\, dx = 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)} \cos{\left(3 x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{2}{\left(x \right)} \cos{\left(x \right)} \cos{\left(3 x \right)} = 4 \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)} - 3 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 4 \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 4 \int \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Rewrite the integrand:
      
      $\sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(- \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} + \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin^{3}{\left(u \right)}}{48} - \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du = \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, du = \frac{u}{16}$
      
      The result is: $\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} + \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}$
      
      So, the result is: $\frac{x}{4} + \frac{\sin^{3}{\left(2 x \right)}}{12} - \frac{\sin{\left(4 x \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 3 \int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Rewrite the integrand:
      
      $\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, du = \frac{u}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      The result is: $\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
      
      So, the result is: $- \frac{3 x}{8} + \frac{3 \sin{\left(4 x \right)}}{32}$
    The result is: $- \frac{x}{8} + \frac{\sin^{3}{\left(2 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{32}$
  So, the result is: $- \frac{x}{4} + \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(4 x \right)}}{16}$
Method #2
1. Rewrite the integrand:
  
  $\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(3 x \right)} = 8 \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)} - 6 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 8 \int \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2}$
    2. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(- \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} + \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin^{3}{\left(u \right)}}{48} - \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du = \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, du = \frac{u}{16}$
      
      The result is: $\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} + \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}$
    So, the result is: $\frac{x}{2} + \frac{\sin^{3}{\left(2 x \right)}}{6} - \frac{\sin{\left(4 x \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 6 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 6 \int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
    2. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, du = \frac{u}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      The result is: $\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
    So, the result is: $- \frac{3 x}{4} + \frac{3 \sin{\left(4 x \right)}}{16}$
  The result is: $- \frac{x}{4} + \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(4 x \right)}}{16}$
Add the constant of integration:

$- \frac{x}{4} + \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(4 x \right)}}{16}+ \mathrm{constant}$

The answer is:

$- \frac{x}{4} + \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(4 x \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                         3                
 |                                   x   sin (2*x)   sin(4*x)
 | sin(2*x)*sin(x)*cos(3*x) dx = C - - + --------- + --------
 |                                   4       6          16   
/

$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(3 x \right)}\, dx = C - \frac{x}{4} + \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(4 x \right)}}{16}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  cos(1)*cos(2)*cos(3)   cos(1)*sin(2)*sin(3)   cos(2)*sin(1)*sin(3)   cos(1)*cos(3)*sin(2)   cos(3)*sin(1)*sin(2)   cos(1)*cos(2)*sin(3)   5*sin(1)*sin(2)*sin(3)
- -------------------- - -------------------- - -------------------- - -------------------- + -------------------- + -------------------- + ----------------------
           4                      4                      4                      8                      4                      6                       24

$$\frac{\sin{\left(1 \right)} \sin{\left(2 \right)} \cos{\left(3 \right)}}{4} - \frac{\cos{\left(1 \right)} \cos{\left(2 \right)} \cos{\left(3 \right)}}{4} - \frac{\sin{\left(2 \right)} \sin{\left(3 \right)} \cos{\left(1 \right)}}{4} + \frac{\sin{\left(3 \right)} \cos{\left(1 \right)} \cos{\left(2 \right)}}{6} - \frac{\sin{\left(1 \right)} \sin{\left(3 \right)} \cos{\left(2 \right)}}{4} + \frac{5 \sin{\left(1 \right)} \sin{\left(2 \right)} \sin{\left(3 \right)}}{24} - \frac{\sin{\left(2 \right)} \cos{\left(1 \right)} \cos{\left(3 \right)}}{8}$$

=

  cos(1)*cos(2)*cos(3)   cos(1)*sin(2)*sin(3)   cos(2)*sin(1)*sin(3)   cos(1)*cos(3)*sin(2)   cos(3)*sin(1)*sin(2)   cos(1)*cos(2)*sin(3)   5*sin(1)*sin(2)*sin(3)
- -------------------- - -------------------- - -------------------- - -------------------- + -------------------- + -------------------- + ----------------------
           4                      4                      4                      8                      4                      6                       24

$$\frac{\sin{\left(1 \right)} \sin{\left(2 \right)} \cos{\left(3 \right)}}{4} - \frac{\cos{\left(1 \right)} \cos{\left(2 \right)} \cos{\left(3 \right)}}{4} - \frac{\sin{\left(2 \right)} \sin{\left(3 \right)} \cos{\left(1 \right)}}{4} + \frac{\sin{\left(3 \right)} \cos{\left(1 \right)} \cos{\left(2 \right)}}{6} - \frac{\sin{\left(1 \right)} \sin{\left(3 \right)} \cos{\left(2 \right)}}{4} + \frac{5 \sin{\left(1 \right)} \sin{\left(2 \right)} \sin{\left(3 \right)}}{24} - \frac{\sin{\left(2 \right)} \cos{\left(1 \right)} \cos{\left(3 \right)}}{8}$$

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Numerical answer [src]

-0.17199566517858

-0.17199566517858

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Identical expressions

Integral of sin2xsinxcos3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2