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Integral of d{x}:
Integral of sin(x²)
Integral of 1/(1+x^3)^(1/3)
Integral of x/(x^2-x)
Integral of x^2cos3xdx
Identical expressions
x^2cos3xdx
x squared co sinus of e of 3xdx
x2cos3xdx
x²cos3xdx
x to the power of 2cos3xdx

Solving integrals/ x^2cos3xdx

Integral of x^2cos3xdx dx

The solution

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  1                 
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 |   2              
 |  x *cos(3*x)*1 dx
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0

$$\int\limits_{0}^{1} x^{2} \cos{\left(3 x \right)} 1\, dx$$

Integral(x^2*cos(3*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 $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .

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

To find $v{\left(x \right)}$ :
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
  
  $\int \frac{\cos{\left(u \right)}}{9}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
    So, the result is: $\frac{\sin{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $\frac{\sin{\left(3 x \right)}}{3}$
Now evaluate the sub-integral.
Use integration by parts:

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

Let $u{\left(x \right)} = \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = \frac{2}{3}$ .

To find $v{\left(x \right)}$ :
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
  
  $\int \frac{\sin{\left(u \right)}}{9}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $- \frac{\cos{\left(3 x \right)}}{3}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \left(- \frac{2 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}$
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
  
  $\int \frac{\cos{\left(u \right)}}{9}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
    So, the result is: $\frac{\sin{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $\frac{\sin{\left(3 x \right)}}{3}$
So, the result is: $- \frac{2 \sin{\left(3 x \right)}}{27}$
Add the constant of integration:

$\frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{2 x \cos{\left(3 x \right)}}{9} - \frac{2 \sin{\left(3 x \right)}}{27}+ \mathrm{constant}$

The answer is:

$\frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{2 x \cos{\left(3 x \right)}}{9} - \frac{2 \sin{\left(3 x \right)}}{27}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                              
 |                                      2                        
 |  2                     2*sin(3*x)   x *sin(3*x)   2*x*cos(3*x)
 | x *cos(3*x)*1 dx = C - ---------- + ----------- + ------------
 |                            27            3             9      
/

$${{\left(9\,x^2-2\right)\,\sin \left(3\,x\right)+6\,x\,\cos \left(3 \,x\right)}\over{27}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

2*cos(3)   7*sin(3)
-------- + --------
   9          27

$${{7\,\sin 3+6\,\cos 3}\over{27}}$$

2*cos(3)   7*sin(3)
-------- + --------
   9          27

$$\frac{2 \cos{\left(3 \right)}}{9} + \frac{7 \sin{\left(3 \right)}}{27}$$

Simplify

Numerical answer [src]

-0.183411663821615

-0.183411663821615

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^2cos3xdx dx

Limits of integration:

The graph:

Piecewise:

The solution