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Identical expressions
x×sin4x^2dx
x× sinus of 4x squared dx
x×sin4x2dx
x×sin4x²dx
x×sin4x to the power of 2dx

Solving integrals/ x×sin4x^2dx

Integral of x×sin4x^2dx dx

The solution

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  1                 
  /                 
 |                  
 |       2          
 |  x*sin (4*x)*1 dx
 |                  
/                   
0

$$\int\limits_{0}^{1} x \sin^{2}{\left(4 x \right)} 1\, dx$$

Integral(x*sin(4*x)^2*1, (x, 0, 1))

Detail solution

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 $\operatorname{dv}{\left(x \right)} = \sin^{2}{\left(4 x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = 1$ .

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\sin^{2}{\left(4 x \right)} = \frac{1}{2} - \frac{\cos{\left(8 x \right)}}{2}$
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:
    
    $\int \left(- \frac{\cos{\left(8 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(8 x \right)}\, dx}{2}$
    1. Let $u = 8 x$ .
      
      Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{\cos{\left(u \right)}}{64}\, du$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
        
        The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
        
        So, the result is: $\frac{\sin{\left(u \right)}}{8}$
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(8 x \right)}}{8}$
    So, the result is: $- \frac{\sin{\left(8 x \right)}}{16}$
  The result is: $\frac{x}{2} - \frac{\sin{\left(8 x \right)}}{16}$
Now evaluate the sub-integral.
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:
  
  $\int \left(- \frac{\sin{\left(8 x \right)}}{16}\right)\, dx = - \frac{\int \sin{\left(8 x \right)}\, dx}{16}$
  1. Let $u = 8 x$ .
    
    Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
    
    $\int \frac{\sin{\left(u \right)}}{64}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{8}\, du = \frac{\int \sin{\left(u \right)}\, du}{8}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $- \frac{\cos{\left(u \right)}}{8}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(8 x \right)}}{8}$
  So, the result is: $\frac{\cos{\left(8 x \right)}}{128}$
The result is: $\frac{x^{2}}{4} + \frac{\cos{\left(8 x \right)}}{128}$
Now simplify:

$\frac{x^{2}}{4} - \frac{x \sin{\left(8 x \right)}}{16} - \frac{\cos{\left(8 x \right)}}{128}$
Add the constant of integration:

$\frac{x^{2}}{4} - \frac{x \sin{\left(8 x \right)}}{16} - \frac{\cos{\left(8 x \right)}}{128}+ \mathrm{constant}$

The answer is:

$\frac{x^{2}}{4} - \frac{x \sin{\left(8 x \right)}}{16} - \frac{\cos{\left(8 x \right)}}{128}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                       
 |                         2                              
 |      2                 x    cos(8*x)     /x   sin(8*x)\
 | x*sin (4*x)*1 dx = C - -- - -------- + x*|- - --------|
 |                        4      128        \2      16   /
/

$$-{{8\,x\,\sin \left(8\,x\right)+\cos \left(8\,x\right)-32\,x^2 }\over{128}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   2            2                   
cos (4)   17*sin (4)   cos(4)*sin(4)
------- + ---------- - -------------
   4          64             8

$${{1}\over{128}}-{{8\,\sin 8+\cos 8-32}\over{128}}$$

   2            2                   
cos (4)   17*sin (4)   cos(4)*sin(4)
------- + ---------- - -------------
   4          64             8

$$- \frac{\sin{\left(4 \right)} \cos{\left(4 \right)}}{8} + \frac{\cos^{2}{\left(4 \right)}}{4} + \frac{17 \sin^{2}{\left(4 \right)}}{64}$$

Упростить

Numerical answer [src]

0.197114328600168

0.197114328600168

The graph

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

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Identical expressions

Integral of x×sin4x^2dx dx

Limits of integration:

The graph:

Piecewise:

The solution