Integral of sin3xsinx dx

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

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

Detail solution

Rewrite the integrand:

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

$\sin^{3}{\left(x \right)} \cos{\left(x \right)}$
Add the constant of integration:

$\sin^{3}{\left(x \right)} \cos{\left(x \right)}+ \mathrm{constant}$

The answer is:

$\sin^{3}{\left(x \right)} \cos{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                          sin(4*x)   sin(2*x)
 | sin(3*x)*sin(x) dx = C - -------- + --------
 |                             8          4    
/

$$\int \sin{\left(x \right)} \sin{\left(3 x \right)}\, dx = C + \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{8}$$

Simplify

The graph

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The answer [src]

  3*cos(3)*sin(1)   cos(1)*sin(3)
- --------------- + -------------
         8                8

$$\frac{\sin{\left(3 \right)} \cos{\left(1 \right)}}{8} - \frac{3 \sin{\left(1 \right)} \cos{\left(3 \right)}}{8}$$

=

  3*cos(3)*sin(1)   cos(1)*sin(3)
- --------------- + -------------
         8                8

$$\frac{\sin{\left(3 \right)} \cos{\left(1 \right)}}{8} - \frac{3 \sin{\left(1 \right)} \cos{\left(3 \right)}}{8}$$

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Numerical answer [src]

0.321924668619911

0.321924668619911

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Integral of sin3xsinx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2