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xsinx^2dx

Integral of x*sinx^2dx dx

The solution

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

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

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

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:
    
    $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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.
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
    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
    1. 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: $\frac{x^{2}}{4} + \frac{\cos{\left(2 x \right)}}{8}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The answer [src]

       2                   
1   sin (1)   cos(1)*sin(1)
- + ------- - -------------
4      4            2

$${{1}\over{8}}-{{2\,\sin 2+\cos 2-2}\over{8}}$$

       2                   
1   sin (1)   cos(1)*sin(1)
- + ------- - -------------
4      4            2

$$- \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{2} + \frac{\sin^{2}{\left(1 \right)}}{4} + \frac{1}{4}$$

Упростить

Numerical answer [src]

0.199693997861972

0.199693997861972

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

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Integral of x*sinx^2dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2