Integral of ∫sin²xcos²xdx dx

The solution

You have entered [src]

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

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

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

Detail solution

Rewrite the integrand:

$\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} 1 = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x$ .
  
  Then let $du = 2 dx$ and substitute $du$ :
  
  $\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, du = \frac{u}{8}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
    The result is: $\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
  Now substitute $u$ back in:
  
  $\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(2 x \right)}}{4}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{4}\, dx = \frac{x}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
    So, the result is: $- \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
  The result is: $\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
Method #3
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(2 x \right)}}{4}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{4}\, dx = \frac{x}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
    So, the result is: $- \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
  The result is: $\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
Add the constant of integration:

$\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer is:

$\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                       
 |                                        
 |    2       2               sin(4*x)   x
 | sin (x)*cos (x)*1 dx = C - -------- + -
 |                               32      8
/

$$\int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} 1\, dx = C + \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   cos(2)*sin(2)
- - -------------
8         16

$$- \frac{\sin{\left(2 \right)} \cos{\left(2 \right)}}{16} + \frac{1}{8}$$

1   cos(2)*sin(2)
- - -------------
8         16

$$- \frac{\sin{\left(2 \right)} \cos{\left(2 \right)}}{16} + \frac{1}{8}$$

Упростить

Numerical answer [src]

0.148650077978373

0.148650077978373

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of ∫sin²xcos²xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3