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Similar expressions
(4xsin^2x-xcos^2(2x)*2)
What you mean?
4*x*sin(x)^2 + x*cos(2*x)^(2^2)

Solving integrals/ (4xsin^2x+xcos^2(2x)*2)

You entered:

(4xsin^2x+xcos^2(2x)*2)

What you mean?

4*x*sin(x)^2 + x*cos(2*x)^(2^2)

Integral of (4xsin^2x+xcos^2(2x)*2) dx

The solution

You have entered [src]

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

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

Simplify

Detail solution

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

$\frac{3 x^{2}}{2} - x \sin{\left(2 x \right)} + \frac{x \sin{\left(4 x \right)}}{4} + \frac{\cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{2} - \frac{1}{16}$
Add the constant of integration:

$\frac{3 x^{2}}{2} - x \sin{\left(2 x \right)} + \frac{x \sin{\left(4 x \right)}}{4} + \frac{\cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{2} - \frac{1}{16}+ \mathrm{constant}$

The answer is:

$\frac{3 x^{2}}{2} - x \sin{\left(2 x \right)} + \frac{x \sin{\left(4 x \right)}}{4} + \frac{\cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{2} - \frac{1}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                           
 |                                           2                                                                
 | /       2           2       \          3*x    cos(2*x)   cos(4*x)       /x   sin(4*x)\       /x   sin(2*x)\
 | \4*x*sin (x) + x*cos (2*x)*2/ dx = C - ---- - -------- + -------- + 2*x*|- + --------| + 4*x*|- - --------|
 |                                         2        2          16          \2      8    /       \2      4    /
/

$${{4\,x\,\sin \left(4\,x\right)+\cos \left(4\,x\right)+8\,x^2}\over{ 16}}-{{2\,x\,\sin \left(2\,x\right)+\cos \left(2\,x\right)-2\,x^2 }\over{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                                 2            2                                       
         2           2      9*cos (6)   35*sin (6)                     3*cos(6)*sin(6)
1 + 8*cos (3) + 9*sin (3) + --------- + ---------- - 6*cos(3)*sin(3) + ---------------
                                2           8                                 2

$${{12\,\sin 12+\cos 12-48\,\sin 6-8\,\cos 6+223}\over{16}}$$

                                 2            2                                       
         2           2      9*cos (6)   35*sin (6)                     3*cos(6)*sin(6)
1 + 8*cos (3) + 9*sin (3) + --------- + ---------- - 6*cos(3)*sin(3) + ---------------
                                2           8                                 2

$$\frac{3 \sin{\left(6 \right)} \cos{\left(6 \right)}}{2} + 9 \sin^{2}{\left(3 \right)} + \frac{35 \sin^{2}{\left(6 \right)}}{8} - 6 \sin{\left(3 \right)} \cos{\left(3 \right)} + 1 + \frac{9 \cos^{2}{\left(6 \right)}}{2} + 8 \cos^{2}{\left(3 \right)}$$

Simplify

Numerical answer [src]

13.945972535192

13.945972535192

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

What you mean?

You entered:

What you mean?

Integral of (4xsin^2x+xcos^2(2x)*2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2