Mister Exam

Other calculators


2*sin(2*x)

Integral of 2*sin(2*x) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
  1              
  /              
 |               
 |  2*sin(2*x) dx
 |               
/                
0                
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=2sin(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)2du\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=2sin(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:

              udu=udu\int u\, du = - \int u\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

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

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

            Now substitute uu back in:

            cos2(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:

            udu\int u\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

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

            Now substitute uu back in:

            sin2(x)2\frac{\sin^{2}{\left(x \right)}}{2}

        So, the result is: cos2(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]
  /                            
 |                             
 | 2*sin(2*x) dx = C - cos(2*x)
 |                             
/                              
2sin(2x)dx=Ccos(2x)\int 2 \sin{\left(2 x \right)}\, dx = C - \cos{\left(2 x \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
The answer [src]
1 - cos(2)
1cos(2)1 - \cos{\left(2 \right)}
=
=
1 - cos(2)
1cos(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.