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Solving integrals/ -sin2x

Integral of -sin2x dx

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The solution

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 2*pi            
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  |  -sin(2*x) dx
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∫02π(−sin⁡(2x)) dx\int\limits_{0}^{2 \pi} \left(- \sin{\left(2 x \right)}\right)\, dx 
Integral(-sin(2*x), (x, 0, 2*pi))
Detail solution

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

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

    1. There are multiple ways to do this integral.

      Method #1

      1. Let u=2xu = 2 x.

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

      Method #2

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

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

        1. There are multiple ways to do this integral.

          Method #1

          1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

            Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

            ∫(−u) du\int \left(- u\right)\, du

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

              ∫u du=−∫u du\int u\, du = - \int u\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                ∫u du=u22\int u\, du = \frac{u^{2}}{2}

              So, the result is: −u22- \frac{u^{2}}{2}

            Now substitute uu back in:

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

          Method #2

          1. Let u=sin⁡(x)u = \sin{\left(x \right)}.

            Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

            ∫u du\int u\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫u du=u22\int u\, du = \frac{u^{2}}{2}

            Now substitute uu back in:

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

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

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

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                           
 |                    cos(2*x)
 | -sin(2*x) dx = C + --------
 |                       2    
/
∫(−sin⁡(2x)) dx=C+cos⁡(2x)2\int \left(- \sin{\left(2 x \right)}\right)\, dx = C + \frac{\cos{\left(2 x \right)}}{2}
Simplify
The graph
0.00.51.01.52.02.53.03.54.04.55.05.56.02-2
Plot the graph f(x)
The answer [src] 
0
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Numerical answer [src] 
5.4435142372695e-22
5.4435142372695e-22

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

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