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2*sin(2*x)
Solving integrals/ 2*sin(2*x)

Integral of 2*sin(2*x) dx

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∫012sin⁡(2x) dx\int\limits_{0}^{1} 2 \sin{\left(2 x \right)}\, dx 
Integral(2*sin(2*x), (x, 0, 1))
Detail solution

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

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

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
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 | 2*sin(2*x) dx = C - cos(2*x)
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∫2sin⁡(2x) dx=C−cos⁡(2x)\int 2 \sin{\left(2 x \right)}\, dx = C - \cos{\left(2 x \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
1 - cos(2)
1−cos⁡(2)1 - \cos{\left(2 \right)} 
=
= 
1 - cos(2)
1−cos⁡(2)1 - \cos{\left(2 \right)} 
1 - cos(2)
Numerical answer [src] 
1.41614683654714
1.41614683654714
The graph
Integral of 2*sin(2*x) dx

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

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