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(x^ two)sin4xdx
(x squared ) sinus of 4xdx
(x to the power of two) sinus of 4xdx
(x2)sin4xdx
x2sin4xdx
(x²)sin4xdx
(x to the power of 2)sin4xdx
x^2sin4xdx

Solving integrals/ (x^2)sin4xdx

Integral of (x^2)sin4xdx dx

The solution

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

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

Integral(x^2*sin(4*x)*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 $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = 2 x$ .

To find $v{\left(x \right)}$ :
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{\sin{\left(u \right)}}{16}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
  Now substitute $u$ back in:
  
  $- \frac{\cos{\left(4 x \right)}}{4}$
Now evaluate the sub-integral.
Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = - \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = - \frac{1}{2}$ .

To find $v{\left(x \right)}$ :
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{\cos{\left(u \right)}}{16}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
    1. 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:
  
  $\frac{\sin{\left(4 x \right)}}{4}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}$
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{\sin{\left(u \right)}}{16}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
  Now substitute $u$ back in:
  
  $- \frac{\cos{\left(4 x \right)}}{4}$
So, the result is: $\frac{\cos{\left(4 x \right)}}{32}$
Add the constant of integration:

$- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{\cos{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer is:

$- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{\cos{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                          
 |                                    2                      
 |  2                     cos(4*x)   x *cos(4*x)   x*sin(4*x)
 | x *sin(4*x)*1 dx = C + -------- - ----------- + ----------
 |                           32           4            8     
/

$${{8\,x\,\sin \left(4\,x\right)+\left(2-16\,x^2\right)\,\cos \left(4 \,x\right)}\over{64}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  1    7*cos(4)   sin(4)
- -- - -------- + ------
  32      32        8

$${{4\,\sin 4-7\,\cos 4}\over{32}}-{{1}\over{32}}$$

  1    7*cos(4)   sin(4)
- -- - -------- + ------
  32      32        8

$$\frac{\sin{\left(4 \right)}}{8} - \frac{1}{32} - \frac{7 \cos{\left(4 \right)}}{32}$$

Упростить

Numerical answer [src]

0.0171342301504241

0.0171342301504241

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 (x^2)sin4xdx dx

Limits of integration:

The graph:

Piecewise:

The solution