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Integral of d{x}:
Integral of xsinx/cos^3x
Integral of sin(3x)^4
Integral of ydx
Integral of 1/y^3
Derivative of:
sin(3x)^4
Identical expressions
sin(3x)^ four
sinus of (3x) to the power of 4
sinus of (3x) to the power of four
sin(3x)4
sin3x4
sin(3x)⁴
sin3x^4
sin(3x)^4dx

Solving integrals/ sin(3x)^4

Integral of sin(3x)^4 dx

The solution

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  1             
  /             
 |              
 |     4        
 |  sin (3*x) dx
 |              
/               
0

$$\int\limits_{0}^{1} \sin^{4}{\left(3 x \right)}\, dx$$

Integral(sin(3*x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(3 x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2}$
There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(6 x \right)}}{4} - \frac{\cos{\left(6 x \right)}}{2} + \frac{1}{4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{2}{\left(6 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(6 x \right)} = \frac{\cos{\left(12 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(12 x \right)}}{2}\, dx = \frac{\int \cos{\left(12 x \right)}\, dx}{2}$
      
      Let $u = 12 x$ .
      
      Then let $du = 12 dx$ and substitute $\frac{du}{12}$ :
      
      $\int \frac{\cos{\left(u \right)}}{144}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{12}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{12}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(12 x \right)}}{12}$
      
      So, the result is: $\frac{\sin{\left(12 x \right)}}{24}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(12 x \right)}}{24}$
    So, the result is: $\frac{x}{8} + \frac{\sin{\left(12 x \right)}}{96}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(6 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{2}$
    1. Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
    So, the result is: $- \frac{\sin{\left(6 x \right)}}{12}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{4}\, dx = \frac{x}{4}$
  The result is: $\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(6 x \right)}}{4} - \frac{\cos{\left(6 x \right)}}{2} + \frac{1}{4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{2}{\left(6 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(6 x \right)} = \frac{\cos{\left(12 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(12 x \right)}}{2}\, dx = \frac{\int \cos{\left(12 x \right)}\, dx}{2}$
      
      Let $u = 12 x$ .
      
      Then let $du = 12 dx$ and substitute $\frac{du}{12}$ :
      
      $\int \frac{\cos{\left(u \right)}}{144}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{12}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{12}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(12 x \right)}}{12}$
      
      So, the result is: $\frac{\sin{\left(12 x \right)}}{24}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(12 x \right)}}{24}$
    So, the result is: $\frac{x}{8} + \frac{\sin{\left(12 x \right)}}{96}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(6 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{2}$
    1. Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
    So, the result is: $- \frac{\sin{\left(6 x \right)}}{12}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{4}\, dx = \frac{x}{4}$
  The result is: $\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}$
Add the constant of integration:

$\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}+ \mathrm{constant}$

The answer is:

$\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                                              
 |    4               sin(6*x)   sin(12*x)   3*x
 | sin (3*x) dx = C - -------- + --------- + ---
 |                       12          96       8 
/

$${{{{{{\sin \left(12\,x\right)}\over{2}}+6\,x}\over{8}}-{{\sin \left(6\,x\right)}\over{2}}+{{3\,x}\over{2}}}\over{6}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                       3          
3   cos(3)*sin(3)   sin (3)*cos(3)
- - ------------- - --------------
8         8               12

$${{\sin 12-8\,\sin 6+36}\over{96}}$$

                       3          
3   cos(3)*sin(3)   sin (3)*cos(3)
- - ------------- - --------------
8         8               12

$$- \frac{\sin^{3}{\left(3 \right)} \cos{\left(3 \right)}}{12} - \frac{\sin{\left(3 \right)} \cos{\left(3 \right)}}{8} + \frac{3}{8}$$

Simplify

Numerical answer [src]

0.392695323620739

0.392695323620739

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin(3x)^4 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2