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(1-cos2x)/2
Solving integrals/ (1-cos2x)/2

Integral of (1-cos2x)/2 dx

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The solution

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  1                
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 |  1 - cos(2*x)   
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∫011−cos⁡(2x)2 dx\int\limits_{0}^{1} \frac{1 - \cos{\left(2 x \right)}}{2}\, dx 
Integral((1 - cos(2*x))/2, (x, 0, 1))
Detail solution

  1. The integral of a constant times a function is the constant times the integral of the function:

    ∫1−cos⁡(2x)2 dx=∫(1−cos⁡(2x)) dx2\int \frac{1 - \cos{\left(2 x \right)}}{2}\, dx = \frac{\int \left(1 - \cos{\left(2 x \right)}\right)\, dx}{2}

    1. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

        ∫1 dx=x\int 1\, dx = x

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫(−cos⁡(2x)) dx=−∫cos⁡(2x) dx\int \left(- \cos{\left(2 x \right)}\right)\, dx = - \int \cos{\left(2 x \right)}\, dx

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

          1. The integral of a constant times a function is the constant times the integral of the function:

            ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            1. The integral of cosine is sine:

              ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

            So, the result is: sin⁡(u)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡(2x)2- \frac{\sin{\left(2 x \right)}}{2}

      The result is: x−sin⁡(2x)2x - \frac{\sin{\left(2 x \right)}}{2}

    So, the result is: x2−sin⁡(2x)4\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}

  2. Add the constant of integration:

    x2−sin⁡(2x)4+constant\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}+ \mathrm{constant}

The answer is:

x2−sin⁡(2x)4+constant\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                  
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 | 1 - cos(2*x)          x   sin(2*x)
 | ------------ dx = C + - - --------
 |      2                2      4    
 |                                   
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∫1−cos⁡(2x)2 dx=C+x2−sin⁡(2x)4\int \frac{1 - \cos{\left(2 x \right)}}{2}\, dx = C + \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.01.0
Plot the graph f(x)
The answer [src] 
1   sin(2)
- - ------
2     4
12−sin⁡(2)4\frac{1}{2} - \frac{\sin{\left(2 \right)}}{4} 
=
= 
1   sin(2)
- - ------
2     4
12−sin⁡(2)4\frac{1}{2} - \frac{\sin{\left(2 \right)}}{4} 
1/2 - sin(2)/4
Numerical answer [src] 
0.27267564329358
0.27267564329358
The graph
Integral of (1-cos2x)/2 dx

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

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