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Similar expressions
(3+sin(2*x))^2

Integral of (3-sin(2*x))^2 dx

The solution

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 |                2   
 |  (3 - sin(2*x))  dx
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2

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

Integral((3 - sin(2*x))^2, (x, 2, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(3 - \sin{\left(2 x \right)}\right)^{2} = \sin^{2}{\left(2 x \right)} - 6 \sin{\left(2 x \right)} + 9$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\sin^{2}{\left(2 x \right)} = \frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}$
  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:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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 result is: $\frac{x}{2} - \frac{\sin{\left(4 x \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 6 \sin{\left(2 x \right)}\right)\, dx = - 6 \int \sin{\left(2 x \right)}\, dx$
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
      
      Method #2
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
      
      Method #2
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{2}{\left(x \right)}}{2}$
      
      So, the result is: $- \cos^{2}{\left(x \right)}$
    So, the result is: $3 \cos{\left(2 x \right)}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 9\, dx = 9 x$
  The result is: $\frac{19 x}{2} - \frac{\sin{\left(4 x \right)}}{8} + 3 \cos{\left(2 x \right)}$
Method #2
1. Rewrite the integrand:
  
  $\left(3 - \sin{\left(2 x \right)}\right)^{2} = \sin^{2}{\left(2 x \right)} - 6 \sin{\left(2 x \right)} + 9$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\sin^{2}{\left(2 x \right)} = \frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}$
  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:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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 result is: $\frac{x}{2} - \frac{\sin{\left(4 x \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 6 \sin{\left(2 x \right)}\right)\, dx = - 6 \int \sin{\left(2 x \right)}\, dx$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
    So, the result is: $3 \cos{\left(2 x \right)}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 9\, dx = 9 x$
  The result is: $\frac{19 x}{2} - \frac{\sin{\left(4 x \right)}}{8} + 3 \cos{\left(2 x \right)}$
Add the constant of integration:

$\frac{19 x}{2} - \frac{\sin{\left(4 x \right)}}{8} + 3 \cos{\left(2 x \right)}+ \mathrm{constant}$

The answer is:

$\frac{19 x}{2} - \frac{\sin{\left(4 x \right)}}{8} + 3 \cos{\left(2 x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     
 |                                                      
 |               2                       sin(4*x)   19*x
 | (3 - sin(2*x))  dx = C + 3*cos(2*x) - -------- + ----
 |                                          8        2  
/

$$\int \left(3 - \sin{\left(2 x \right)}\right)^{2}\, dx = C + \frac{19 x}{2} - \frac{\sin{\left(4 x \right)}}{8} + 3 \cos{\left(2 x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

        2         2                                                                             
     cos (2)   sin (2)      2         2                            cos(2)*sin(2)   cos(4)*sin(4)
-9 + ------- + ------- - cos (4) - sin (4) - 3*cos(4) + 3*cos(2) - ------------- + -------------
        2         2                                                      4               4

$$-9 + 3 \cos{\left(2 \right)} - \sin^{2}{\left(4 \right)} - \cos^{2}{\left(4 \right)} + \frac{\cos^{2}{\left(2 \right)}}{2} - \frac{\sin{\left(2 \right)} \cos{\left(2 \right)}}{4} + \frac{\sin{\left(4 \right)} \cos{\left(4 \right)}}{4} + \frac{\sin^{2}{\left(2 \right)}}{2} - 3 \cos{\left(4 \right)}$$

        2         2                                                                             
     cos (2)   sin (2)      2         2                            cos(2)*sin(2)   cos(4)*sin(4)
-9 + ------- + ------- - cos (4) - sin (4) - 3*cos(4) + 3*cos(2) - ------------- + -------------
        2         2                                                      4               4

-9 + cos(2)^2/2 + sin(2)^2/2 - cos(4)^2 - sin(4)^2 - 3*cos(4) + 3*cos(2) - cos(2)*sin(2)/4 + cos(4)*sin(4)/4

Numerical answer [src]

-8.56923955430918

-8.56923955430918

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

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How to use it?

Identical expressions

Similar expressions

Integral of (3-sin(2*x))^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #1

Method #2

Method #2