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Solving integrals/ (sin(2x-1))*dx

Integral of (sin(2x-1))*dx dx

Limits of integration:

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The solution

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

  1. Let u=2x−1u = 2 x - 1.

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

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

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

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

      1. The integral of sine is negative cosine:

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

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

    Now substitute uu back in:

    −cos⁡(2x−1)2- \frac{\cos{\left(2 x - 1 \right)}}{2}

  2. Now simplify:

    −cos⁡(2x−1)2- \frac{\cos{\left(2 x - 1 \right)}}{2}

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                  
 |                       cos(2*x - 1)
 | sin(2*x - 1) dx = C - ------------
 |                            2      
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∫sin⁡(2x−1) dx=C−cos⁡(2x−1)2\int \sin{\left(2 x - 1 \right)}\, dx = C - \frac{\cos{\left(2 x - 1 \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
Plot the graph f(x)
The answer [src] 
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Numerical answer [src] 
1.28230080812608e-23
1.28230080812608e-23

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

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