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cos^2(3x)sin^4(3x)dx

Solving integrals/ cos^2(3x)sin^4(3x)

Integral of cos^2(3x)sin^4(3x) dx

The solution

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  0                       
  /                       
 |                        
 |     2         4        
 |  cos (3*x)*sin (3*x) dx
 |                        
/                         
0

$$\int\limits_{0}^{0} \sin^{4}{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\, dx$$

Integral(cos(3*x)^2*sin(3*x)^4, (x, 0, 0))

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(3 x \right)} \cos^{2}{\left(3 x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(6 x \right)}}{2} + \frac{1}{2}\right)$
There are multiple ways to do this integral.
Method #1
1. Let $u = 6 x$ .
  
  Then let $du = 6 dx$ and substitute $du$ :
  
  $\int \left(\frac{\cos^{3}{\left(u \right)}}{48} - \frac{\cos^{2}{\left(u \right)}}{48} - \frac{\cos{\left(u \right)}}{48} + \frac{1}{48}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{48}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{48}$
      
      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: $- \frac{\sin^{3}{\left(u \right)}}{144} + \frac{\sin{\left(u \right)}}{48}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{48}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{48}$
      
      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 $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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: $- \frac{u}{96} - \frac{\sin{\left(2 u \right)}}{192}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(u \right)}}{48}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{48}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin{\left(u \right)}}{48}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{48}\, du = \frac{u}{48}$
    The result is: $\frac{u}{96} - \frac{\sin{\left(2 u \right)}}{192} - \frac{\sin^{3}{\left(u \right)}}{144}$
  Now substitute $u$ back in:
  
  $\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(6 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(6 x \right)}}{8} - \frac{\cos^{2}{\left(6 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{3}{\left(6 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(6 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(6 x \right)} = \left(1 - \sin^{2}{\left(6 x \right)}\right) \cos{\left(6 x \right)}$
    2. Let $u = \sin{\left(6 x \right)}$ .
      
      Then let $du = 6 \cos{\left(6 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{6} - \frac{u^{2}}{6}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{6}\, du = \frac{u}{6}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}$
      
      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: $- \frac{u^{3}}{18}$
      
      The result is: $- \frac{u^{3}}{18} + \frac{u}{6}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(6 x \right)}}{18} + \frac{\sin{\left(6 x \right)}}{6}$
    So, the result is: $- \frac{\sin^{3}{\left(6 x \right)}}{144} + \frac{\sin{\left(6 x \right)}}{48}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{8}$
    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 $\frac{du}{12}$ :
      
      $\int \frac{\cos{\left(u \right)}}{12}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}$
      
      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: $- \frac{x}{16} - \frac{\sin{\left(12 x \right)}}{192}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{8}$
    1. Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      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: $- \frac{\sin{\left(6 x \right)}}{48}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dx = \frac{x}{8}$
  The result is: $\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}$
Method #3
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(6 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(6 x \right)}}{8} - \frac{\cos^{2}{\left(6 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{3}{\left(6 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(6 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(6 x \right)} = \left(1 - \sin^{2}{\left(6 x \right)}\right) \cos{\left(6 x \right)}$
    2. Let $u = \sin{\left(6 x \right)}$ .
      
      Then let $du = 6 \cos{\left(6 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{6} - \frac{u^{2}}{6}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{6}\, du = \frac{u}{6}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}$
      
      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: $- \frac{u^{3}}{18}$
      
      The result is: $- \frac{u^{3}}{18} + \frac{u}{6}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(6 x \right)}}{18} + \frac{\sin{\left(6 x \right)}}{6}$
    So, the result is: $- \frac{\sin^{3}{\left(6 x \right)}}{144} + \frac{\sin{\left(6 x \right)}}{48}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{8}$
    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 $\frac{du}{12}$ :
      
      $\int \frac{\cos{\left(u \right)}}{12}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}$
      
      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: $- \frac{x}{16} - \frac{\sin{\left(12 x \right)}}{192}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{8}$
    1. Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      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: $- \frac{\sin{\left(6 x \right)}}{48}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dx = \frac{x}{8}$
  The result is: $\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}$
Add the constant of integration:

$\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}+ \mathrm{constant}$

The answer is:

$\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                       
 |                                 3                      
 |    2         4               sin (6*x)   sin(12*x)   x 
 | cos (3*x)*sin (3*x) dx = C - --------- - --------- + --
 |                                 144         192      16
/

$$\int \sin^{4}{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\, dx = C + \frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

Numerical answer [src]

0.0

0.0

The graph

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

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Identical expressions

Integral of cos^2(3x)sin^4(3x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3