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Solving integrals/ x^3sinx^2

Integral of x^3sinx^2 dx

The solution

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

Integral(x^3*sin(x)^2, (x, 0, 1))

Detail solution

Use integration by parts:

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

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

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

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.
Use integration by parts:

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

Let $u{\left(x \right)} = \frac{3 x^{2}}{4}$ and let $\operatorname{dv}{\left(x \right)} = 2 x - \sin{\left(2 x \right)}$ .

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

To find $v{\left(x \right)}$ :
1. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x\, dx = 2 \int x\, dx$
    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: $x^{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(2 x \right)}\right)\, dx = - \int \sin{\left(2 x \right)}\, dx$
    1. There are multiple ways to do this integral.
      Method #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
      
      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: $\frac{\cos{\left(2 x \right)}}{2}$
  The result is: $x^{2} + \frac{\cos{\left(2 x \right)}}{2}$
Now evaluate the sub-integral.
Use integration by parts:

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

Let $u{\left(x \right)} = \frac{3 x}{4}$ and let $\operatorname{dv}{\left(x \right)} = 2 x^{2} + \cos{\left(2 x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = \frac{3}{4}$ .

To find $v{\left(x \right)}$ :
1. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x^{2}\, dx = 2 \int x^{2}\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    So, the result is: $\frac{2 x^{3}}{3}$
  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:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      1. The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 x \right)}}{2}$
  The result is: $\frac{2 x^{3}}{3} + \frac{\sin{\left(2 x \right)}}{2}$
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:
  
  $\int \frac{x^{3}}{2}\, dx = \frac{\int x^{3}\, dx}{2}$
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{3}\, dx = \frac{x^{4}}{4}$
  So, the result is: $\frac{x^{4}}{8}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{3 \sin{\left(2 x \right)}}{8}\, dx = \frac{3 \int \sin{\left(2 x \right)}\, dx}{8}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\sin{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(2 x \right)}}{2}$
  So, the result is: $- \frac{3 \cos{\left(2 x \right)}}{16}$
The result is: $\frac{x^{4}}{8} - \frac{3 \cos{\left(2 x \right)}}{16}$
Now simplify:

$\frac{x^{4}}{8} - \frac{x^{3} \sin{\left(2 x \right)}}{4} - \frac{3 x^{2} \cos{\left(2 x \right)}}{8} + \frac{3 x \sin{\left(2 x \right)}}{8} + \frac{3 \cos{\left(2 x \right)}}{16}$
Add the constant of integration:

$\frac{x^{4}}{8} - \frac{x^{3} \sin{\left(2 x \right)}}{4} - \frac{3 x^{2} \cos{\left(2 x \right)}}{8} + \frac{3 x \sin{\left(2 x \right)}}{8} + \frac{3 \cos{\left(2 x \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{x^{4}}{8} - \frac{x^{3} \sin{\left(2 x \right)}}{4} - \frac{3 x^{2} \cos{\left(2 x \right)}}{8} + \frac{3 x \sin{\left(2 x \right)}}{8} + \frac{3 \cos{\left(2 x \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

                                                                                        /              3\
  /                                                             2 / 2   cos(2*x)\       |sin(2*x)   2*x |
 |                      4                                    3*x *|x  + --------|   3*x*|-------- + ----|
 |  3    2             x    3*cos(2*x)    3 /x   sin(2*x)\        \        2    /       \   2        3  /
 | x *sin (x) dx = C - -- + ---------- + x *|- - --------| - -------------------- + ---------------------
 |                     8        16          \2      4    /            4                       4          
/

$$-{{\left(4\,x^3-6\,x\right)\,\sin \left(2\,x\right)+\left(6\,x^2-3 \right)\,\cos \left(2\,x\right)-2\,x^4}\over{16}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     2         2                   
  cos (1)   sin (1)   cos(1)*sin(1)
- ------- + ------- + -------------
     4         8            4

$${{2\,\sin 2-3\,\cos 2+2}\over{16}}-{{3}\over{16}}$$

     2         2                   
  cos (1)   sin (1)   cos(1)*sin(1)
- ------- + ------- + -------------
     4         8            4

$$- \frac{\cos^{2}{\left(1 \right)}}{4} + \frac{\sin^{2}{\left(1 \right)}}{8} + \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{4}$$

Упростить

Numerical answer [src]

0.129189710205799

0.129189710205799

The graph

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

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How to use it?

Identical expressions

Similar expressions

Integral of x^3sinx^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2