Integral of xcos4xdx dx

The solution

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  1                
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 |  x*cos(4*x)*1 dx
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$$\int\limits_{0}^{1} x \cos{\left(4 x \right)} 1\, dx$$

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

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

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:

$\int \frac{\sin{\left(4 x \right)}}{4}\, dx = \frac{\int \sin{\left(4 x \right)}\, dx}{4}$
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: $- \frac{\cos{\left(4 x \right)}}{16}$
Add the constant of integration:

$\frac{x \sin{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{x \sin{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

  1    sin(4)   cos(4)
- -- + ------ + ------
  16     4        16

$${{4\,\sin 4+\cos 4}\over{16}}-{{1}\over{16}}$$

  1    sin(4)   cos(4)
- -- + ------ + ------
  16     4        16

$$\frac{\sin{\left(4 \right)}}{4} - \frac{1}{16} + \frac{\cos{\left(4 \right)}}{16}$$

Упростить

Numerical answer [src]

-0.292553350130958

-0.292553350130958

The graph

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

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Integral of xcos4xdx dx

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