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sin(5x)^ two ×cos(5x)^ two
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sin(5x)^2×cos(5x)^2dx

Solving integrals/ sin(5x)^2×cos(5x)^2

Integral of sin(5x)^2×cos(5x)^2 dx

The solution

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  0                       
  /                       
 |                        
 |     2         2        
 |  sin (5*x)*cos (5*x) dx
 |                        
/                         
0

$$\int\limits_{0}^{0} \sin^{2}{\left(5 x \right)} \cos^{2}{\left(5 x \right)}\, dx$$

Integral(sin(5*x)^2*cos(5*x)^2, (x, 0, 0))

Detail solution

Rewrite the integrand:

$\sin^{2}{\left(5 x \right)} \cos^{2}{\left(5 x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(10 x \right)}}{2}\right) \left(\frac{\cos{\left(10 x \right)}}{2} + \frac{1}{2}\right)$
There are multiple ways to do this integral.
Method #1
1. Let $u = 10 x$ .
  
  Then let $du = 10 dx$ and substitute $du$ :
  
  $\int \left(\frac{1}{40} - \frac{\cos^{2}{\left(u \right)}}{40}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{40}\, du = \frac{u}{40}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{40}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{40}$
      
      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: $- \frac{u}{80} - \frac{\sin{\left(2 u \right)}}{160}$
    The result is: $\frac{u}{80} - \frac{\sin{\left(2 u \right)}}{160}$
  Now substitute $u$ back in:
  
  $\frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(10 x \right)}}{2}\right) \left(\frac{\cos{\left(10 x \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(10 x \right)}}{4}$
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:
    
    $\int \left(- \frac{\cos^{2}{\left(10 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{2}{\left(10 x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(10 x \right)} = \frac{\cos{\left(20 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(20 x \right)}}{2}\, dx = \frac{\int \cos{\left(20 x \right)}\, dx}{2}$
      
      Let $u = 20 x$ .
      
      Then let $du = 20 dx$ and substitute $\frac{du}{20}$ :
      
      $\int \frac{\cos{\left(u \right)}}{400}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{20}\, du = \frac{\int \cos{\left(u \right)}\, du}{20}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{20}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(20 x \right)}}{20}$
      
      So, the result is: $\frac{\sin{\left(20 x \right)}}{40}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(20 x \right)}}{40}$
    So, the result is: $- \frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}$
  The result is: $\frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}$
Method #3
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(10 x \right)}}{2}\right) \left(\frac{\cos{\left(10 x \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(10 x \right)}}{4}$
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:
    
    $\int \left(- \frac{\cos^{2}{\left(10 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{2}{\left(10 x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(10 x \right)} = \frac{\cos{\left(20 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(20 x \right)}}{2}\, dx = \frac{\int \cos{\left(20 x \right)}\, dx}{2}$
      
      Let $u = 20 x$ .
      
      Then let $du = 20 dx$ and substitute $\frac{du}{20}$ :
      
      $\int \frac{\cos{\left(u \right)}}{400}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{20}\, du = \frac{\int \cos{\left(u \right)}\, du}{20}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{20}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(20 x \right)}}{20}$
      
      So, the result is: $\frac{\sin{\left(20 x \right)}}{40}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(20 x \right)}}{40}$
    So, the result is: $- \frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}$
  The result is: $\frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}$
Add the constant of integration:

$\frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}+ \mathrm{constant}$

The answer is:

$\frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                          
 |                                           
 |    2         2               sin(20*x)   x
 | sin (5*x)*cos (5*x) dx = C - --------- + -
 |                                 160      8
/

$$\int \sin^{2}{\left(5 x \right)} \cos^{2}{\left(5 x \right)}\, dx = C + \frac{x}{8} - \frac{\sin{\left(20 x \right)}}{160}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

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Numerical answer [src]

0.0

0.0

The graph

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

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Identical expressions

Integral of sin(5x)^2×cos(5x)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3