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Integral of d{x}:
Integral of x/(sinx^2)^2
Integral of (sqrt(x)-2)^2/x
Integral of (sin(3x))^5
Integral of e^xsin(x/2)
Identical expressions
(sin(3x))^ five
( sinus of (3x)) to the power of 5
( sinus of (3x)) to the power of five
(sin(3x))5
sin3x5
(sin(3x))⁵
sin3x^5
(sin(3x))^5dx

Solving integrals/ (sin(3x))^5

Integral of (sin(3x))^5 dx

The solution

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  1             
  /             
 |              
 |     5        
 |  sin (3*x) dx
 |              
/               
0

$$\int\limits_{0}^{1} \sin^{5}{\left(3 x \right)}\, dx$$

Integral(sin(3*x)^5, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{5}{\left(3 x \right)} = \left(1 - \cos^{2}{\left(3 x \right)}\right)^{2} \sin{\left(3 x \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $du$ :
  
  $\int \left(\frac{\sin{\left(u \right)} \cos^{4}{\left(u \right)}}{3} - \frac{2 \sin{\left(u \right)} \cos^{2}{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{3}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)} \cos^{4}{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)} \cos^{4}{\left(u \right)}\, du}{3}$
      
      Let $u = \cos{\left(u \right)}$ .
      
      Then let $du = - \sin{\left(u \right)} du$ 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:
      
      $- \frac{\cos^{5}{\left(u \right)}}{5}$
      
      So, the result is: $- \frac{\cos^{5}{\left(u \right)}}{15}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2 \sin{\left(u \right)} \cos^{2}{\left(u \right)}}{3}\right)\, du = - \frac{2 \int \sin{\left(u \right)} \cos^{2}{\left(u \right)}\, du}{3}$
      
      Let $u = \cos{\left(u \right)}$ .
      
      Then let $du = - \sin{\left(u \right)} du$ 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:
      
      $- \frac{\cos^{3}{\left(u \right)}}{3}$
      
      So, the result is: $\frac{2 \cos^{3}{\left(u \right)}}{9}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
    The result is: $- \frac{\cos^{5}{\left(u \right)}}{15} + \frac{2 \cos^{3}{\left(u \right)}}{9} - \frac{\cos{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $- \frac{\cos^{5}{\left(3 x \right)}}{15} + \frac{2 \cos^{3}{\left(3 x \right)}}{9} - \frac{\cos{\left(3 x \right)}}{3}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(3 x \right)}\right)^{2} \sin{\left(3 x \right)} = \sin{\left(3 x \right)} \cos^{4}{\left(3 x \right)} - 2 \sin{\left(3 x \right)} \cos^{2}{\left(3 x \right)} + \sin{\left(3 x \right)}$
2. Integrate term-by-term:
  1. Let $u = \cos{\left(3 x \right)}$ .
    
    Then let $du = - 3 \sin{\left(3 x \right)} dx$ and substitute $- \frac{du}{3}$ :
    
    $\int \left(- \frac{u^{4}}{3}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \frac{\int u^{4}\, du}{3}$
      
      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: $- \frac{u^{5}}{15}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{5}{\left(3 x \right)}}{15}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\right)\, dx = - 2 \int \sin{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\, dx$
    1. Let $u = \cos{\left(3 x \right)}$ .
      
      Then let $du = - 3 \sin{\left(3 x \right)} dx$ and substitute $- \frac{du}{3}$ :
      
      $\int \left(- \frac{u^{2}}{3}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = - \frac{\int u^{2}\, du}{3}$
      
      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: $- \frac{u^{3}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(3 x \right)}}{9}$
    So, the result is: $\frac{2 \cos^{3}{\left(3 x \right)}}{9}$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\sin{\left(u \right)}}{3}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(3 x \right)}}{3}$
  The result is: $- \frac{\cos^{5}{\left(3 x \right)}}{15} + \frac{2 \cos^{3}{\left(3 x \right)}}{9} - \frac{\cos{\left(3 x \right)}}{3}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(3 x \right)}\right)^{2} \sin{\left(3 x \right)} = \sin{\left(3 x \right)} \cos^{4}{\left(3 x \right)} - 2 \sin{\left(3 x \right)} \cos^{2}{\left(3 x \right)} + \sin{\left(3 x \right)}$
2. Integrate term-by-term:
  1. Let $u = \cos{\left(3 x \right)}$ .
    
    Then let $du = - 3 \sin{\left(3 x \right)} dx$ and substitute $- \frac{du}{3}$ :
    
    $\int \left(- \frac{u^{4}}{3}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \frac{\int u^{4}\, du}{3}$
      
      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: $- \frac{u^{5}}{15}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{5}{\left(3 x \right)}}{15}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\right)\, dx = - 2 \int \sin{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\, dx$
    1. Let $u = \cos{\left(3 x \right)}$ .
      
      Then let $du = - 3 \sin{\left(3 x \right)} dx$ and substitute $- \frac{du}{3}$ :
      
      $\int \left(- \frac{u^{2}}{3}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = - \frac{\int u^{2}\, du}{3}$
      
      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: $- \frac{u^{3}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(3 x \right)}}{9}$
    So, the result is: $\frac{2 \cos^{3}{\left(3 x \right)}}{9}$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\sin{\left(u \right)}}{3}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(3 x \right)}}{3}$
  The result is: $- \frac{\cos^{5}{\left(3 x \right)}}{15} + \frac{2 \cos^{3}{\left(3 x \right)}}{9} - \frac{\cos{\left(3 x \right)}}{3}$
Now simplify:

$- \frac{\left(3 \sin^{4}{\left(3 x \right)} + 4 \sin^{2}{\left(3 x \right)} + 8\right) \cos{\left(3 x \right)}}{45}$
Add the constant of integration:

$- \frac{\left(3 \sin^{4}{\left(3 x \right)} + 4 \sin^{2}{\left(3 x \right)} + 8\right) \cos{\left(3 x \right)}}{45}+ \mathrm{constant}$

The answer is:

$- \frac{\left(3 \sin^{4}{\left(3 x \right)} + 4 \sin^{2}{\left(3 x \right)} + 8\right) \cos{\left(3 x \right)}}{45}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     
 |                                  5             3     
 |    5               cos(3*x)   cos (3*x)   2*cos (3*x)
 | sin (3*x) dx = C - -------- - --------- + -----------
 |                       3           15           9     
/

$$\int \sin^{5}{\left(3 x \right)}\, dx = C - \frac{\cos^{5}{\left(3 x \right)}}{15} + \frac{2 \cos^{3}{\left(3 x \right)}}{9} - \frac{\cos{\left(3 x \right)}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                 5           3   
8    cos(3)   cos (3)   2*cos (3)
-- - ------ - ------- + ---------
45     3         15         9

$$\frac{2 \cos^{3}{\left(3 \right)}}{9} - \frac{\cos^{5}{\left(3 \right)}}{15} + \frac{8}{45} - \frac{\cos{\left(3 \right)}}{3}$$

                 5           3   
8    cos(3)   cos (3)   2*cos (3)
-- - ------ - ------- + ---------
45     3         15         9

$$\frac{2 \cos^{3}{\left(3 \right)}}{9} - \frac{\cos^{5}{\left(3 \right)}}{15} + \frac{8}{45} - \frac{\cos{\left(3 \right)}}{3}$$

8/45 - cos(3)/3 - cos(3)^5/15 + 2*cos(3)^3/9

Numerical answer [src]

0.355555113446564

0.355555113446564

The graph

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

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How to use it?

Identical expressions

Integral of (sin(3x))^5 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3