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Solving integrals/ 1/1-cos(6x)

Integral of 1/1-cos(6x) dx

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The solution

You have entered [src] 
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 |  (1 - cos(6*x)) dx
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∫p4p6(−cos⁡(6x)+1) dx\int\limits_{\frac{p}{4}}^{\frac{p}{6}} \left(- \cos{\left(6 x \right)} + 1\right)\, dx 
Integral(1 - cos(6*x), (x, p/4, p/6))
Detail solution

  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⁡(6x)) dx=−∫cos⁡(6x) dx\int \left(- \cos{\left(6 x \right)}\right)\, dx = - \int \cos{\left(6 x \right)}\, dx

      1. Let u=6xu = 6 x.

        Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

        ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

          ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

          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)6\frac{\sin{\left(u \right)}}{6}

        Now substitute uu back in:

        sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

      So, the result is: −sin⁡(6x)6- \frac{\sin{\left(6 x \right)}}{6}

    The result is: x−sin⁡(6x)6x - \frac{\sin{\left(6 x \right)}}{6}

  2. Add the constant of integration:

    x−sin⁡(6x)6+constantx - \frac{\sin{\left(6 x \right)}}{6}+ \mathrm{constant}

The answer is:

x−sin⁡(6x)6+constantx - \frac{\sin{\left(6 x \right)}}{6}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                    
 |                             sin(6*x)
 | (1 - cos(6*x)) dx = C + x - --------
 |                                6    
/
x−sin⁡(6 x)6x-{{\sin \left(6\,x\right)}\over{6}}
Simplify
The answer [src] 
                   /3*p\
                sin|---|
  sin(p)   p       \ 2 /
- ------ - -- + --------
    6      12      6
2 sin⁡(3 p2)−2 sin⁡p−p12{{2\,\sin \left({{3\,p}\over{2}}\right)-2\,\sin p-p}\over{12}} 
=
= 
                   /3*p\
                sin|---|
  sin(p)   p       \ 2 /
- ------ - -- + --------
    6      12      6
−p12−sin⁡(p)6+sin⁡(3p2)6- \frac{p}{12} - \frac{\sin{\left(p \right)}}{6} + \frac{\sin{\left(\frac{3 p}{2} \right)}}{6}
Упростить

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

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