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Solving integrals/ x^2cos^2x

Integral of x^2cos^2x dx

The solution

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

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

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

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
    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: $\frac{\sin{\left(2 x \right)}}{4}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{2}\, dx = \frac{x}{2}$
  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 $u{\left(x \right)} = \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = 2 x + \sin{\left(2 x \right)}$ .

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

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. 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: $- \cos^{2}{\left(x \right)}$
  The result is: $x^{2} - \frac{\cos{\left(2 x \right)}}{2}$
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:
  
  $\int \frac{x^{2}}{2}\, dx = \frac{\int x^{2}\, dx}{2}$
  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: $\frac{x^{3}}{6}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}$
  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:
    
    $\frac{\sin{\left(2 x \right)}}{2}$
  So, the result is: $- \frac{\sin{\left(2 x \right)}}{8}$
The result is: $\frac{x^{3}}{6} - \frac{\sin{\left(2 x \right)}}{8}$
Now simplify:

$\frac{x^{3}}{6} + \frac{x^{2} \sin{\left(2 x \right)}}{4} + \frac{x \cos{\left(2 x \right)}}{4} - \frac{\sin{\left(2 x \right)}}{8}$
Add the constant of integration:

$\frac{x^{3}}{6} + \frac{x^{2} \sin{\left(2 x \right)}}{4} + \frac{x \cos{\left(2 x \right)}}{4} - \frac{\sin{\left(2 x \right)}}{8}+ \mathrm{constant}$

The answer is:

\frac{x^{3}}{6} + \frac{x^{2} \sin{\left(2 x \right)}}{4} + \frac{x \cos{\left(2 x \right)}}{4} - \frac{\sin{\left(2 x \right)}}{8}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                                                          / 2   cos(2*x)\
 |                                 3                       x*|x  - --------|
 |  2    2             sin(2*x)   x     2 /x   sin(2*x)\     \        2    /
 | x *cos (x) dx = C - -------- + -- + x *|- + --------| - -----------------
 |                        8       6       \2      4    /           2        
/

{{\left(6\,x^2-3\right)\,\sin \left(2\,x\right)+6\,x\,\cos \left(2 \,x\right)+4\,x^3}\over{24}}

Simplify

The graph

Plot the graph f(x)

The answer [src]

     2           2                   
  sin (1)   5*cos (1)   cos(1)*sin(1)
- ------- + --------- + -------------
     12         12            4

{{3\,\sin 2+6\,\cos 2+4}\over{24}}

     2           2                   
  sin (1)   5*cos (1)   cos(1)*sin(1)
- ------- + --------- + -------------
     12         12            4

- \frac{\sin^{2}{\left(1 \right)}}{12} + \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{4} + \frac{5 \cos^{2}{\left(1 \right)}}{12}

Simplify

Numerical answer [src]

0.176292135883091

0.176292135883091

The graph

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

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Identical expressions

Similar expressions

Integral of x^2cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2