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Solving integrals/ cos(1-2x)dx

Integral of cos(1-2x)dx dx

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The solution

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 |  cos(1 - 2*x)*1 dx
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∫01cos⁡(1−2x)1 dx\int\limits_{0}^{1} \cos{\left(1 - 2 x \right)} 1\, dx 
Integral(cos(1 - 2*x)*1, (x, 0, 1))
Detail solution

  1. Let u=1−2xu = 1 - 2 x.

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

    ∫cos⁡(u)4 du\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:

      ∫(−cos⁡(u)2) du=−∫cos⁡(u) du2\int \left(- \frac{\cos{\left(u \right)}}{2}\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−1)2\frac{\sin{\left(2 x - 1 \right)}}{2}

  2. Add the constant of integration:

    sin⁡(2x−1)2+constant\frac{\sin{\left(2 x - 1 \right)}}{2}+ \mathrm{constant}

The answer is:

sin⁡(2x−1)2+constant\frac{\sin{\left(2 x - 1 \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                     
 |                         sin(-1 + 2*x)
 | cos(1 - 2*x)*1 dx = C + -------------
 |                               2      
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sin⁡(2 x−1)2{{\sin \left(2\,x-1\right)}\over{2}}
Simplify
The answer [src] 
sin(1)
sin⁡1\sin 1 
=
= 
sin(1)
sin⁡(1)\sin{\left(1 \right)}
Упростить
Numerical answer [src] 
0.841470984807897
0.841470984807897

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

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