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Solving integrals/ (x-pi/4)*sin(2*x)

Integral of (x-pi/4)*sin(2*x) dx

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∫0π4(x−π4)sin⁡(2x) dx\int\limits_{0}^{\frac{\pi}{4}} \left(x - \frac{\pi}{4}\right) \sin{\left(2 x \right)}\, dx 
Integral((x - pi/4)*sin(2*x), (x, 0, pi/4))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (x−π4)sin⁡(2x)=xsin⁡(2x)−πsin⁡(2x)4\left(x - \frac{\pi}{4}\right) \sin{\left(2 x \right)} = x \sin{\left(2 x \right)} - \frac{\pi \sin{\left(2 x \right)}}{4}

    2. Integrate term-by-term:

      1. Use integration by parts:

        ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

        Let u(x)=xu{\left(x \right)} = x and let dv⁡(x)=sin⁡(2x)\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}.

        Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

        To find v(x)v{\left(x \right)}:

        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. 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}

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

        Now evaluate the sub-integral.

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

        ∫(−cos⁡(2x)2) dx=−∫cos⁡(2x) dx2\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}

        1. Let u=2xu = 2 x.

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

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

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

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

        So, the result is: −sin⁡(2x)4- \frac{\sin{\left(2 x \right)}}{4}

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

        ∫(−πsin⁡(2x)4) dx=−π∫sin⁡(2x) dx4\int \left(- \frac{\pi \sin{\left(2 x \right)}}{4}\right)\, dx = - \frac{\pi \int \sin{\left(2 x \right)}\, dx}{4}

        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}

        So, the result is: πcos⁡(2x)8\frac{\pi \cos{\left(2 x \right)}}{8}

      The result is: −xcos⁡(2x)2+sin⁡(2x)4+πcos⁡(2x)8- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}

    Method #2

    1. Use integration by parts:

      ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(x)=x−π4u{\left(x \right)} = x - \frac{\pi}{4} and let dv⁡(x)=sin⁡(2x)\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}.

      Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

      To find v(x)v{\left(x \right)}:

      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}

      Now evaluate the sub-integral.

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

      ∫(−cos⁡(2x)2) dx=−∫cos⁡(2x) dx2\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}

      1. Let u=2xu = 2 x.

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

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

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

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

      So, the result is: −sin⁡(2x)4- \frac{\sin{\left(2 x \right)}}{4}

    Method #3

    1. Rewrite the integrand:

      (x−π4)sin⁡(2x)=xsin⁡(2x)−πsin⁡(2x)4\left(x - \frac{\pi}{4}\right) \sin{\left(2 x \right)} = x \sin{\left(2 x \right)} - \frac{\pi \sin{\left(2 x \right)}}{4}

    2. Integrate term-by-term:

      1. Use integration by parts:

        ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

        Let u(x)=xu{\left(x \right)} = x and let dv⁡(x)=sin⁡(2x)\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}.

        Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

        To find v(x)v{\left(x \right)}:

        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}

        Now evaluate the sub-integral.

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

        ∫(−cos⁡(2x)2) dx=−∫cos⁡(2x) dx2\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}

        1. Let u=2xu = 2 x.

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

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

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

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

        So, the result is: −sin⁡(2x)4- \frac{\sin{\left(2 x \right)}}{4}

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

        ∫(−πsin⁡(2x)4) dx=−π∫sin⁡(2x) dx4\int \left(- \frac{\pi \sin{\left(2 x \right)}}{4}\right)\, dx = - \frac{\pi \int \sin{\left(2 x \right)}\, dx}{4}

        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}

        So, the result is: πcos⁡(2x)8\frac{\pi \cos{\left(2 x \right)}}{8}

      The result is: −xcos⁡(2x)2+sin⁡(2x)4+πcos⁡(2x)8- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}

    Method #4

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

      ∫2(x−π4)sin⁡(x)cos⁡(x) dx=2∫(x−π4)sin⁡(x)cos⁡(x) dx\int 2 \left(x - \frac{\pi}{4}\right) \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \left(x - \frac{\pi}{4}\right) \sin{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Rewrite the integrand:

        (x−π4)sin⁡(x)cos⁡(x)=xsin⁡(x)cos⁡(x)−πsin⁡(x)cos⁡(x)4\left(x - \frac{\pi}{4}\right) \sin{\left(x \right)} \cos{\left(x \right)} = x \sin{\left(x \right)} \cos{\left(x \right)} - \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{4}

      2. Integrate term-by-term:

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

          Let u(x)=xu{\left(x \right)} = x and let dv⁡(x)=sin⁡(x)cos⁡(x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}.

          Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

          To find v(x)v{\left(x \right)}:

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

            ∫sin⁡(2x)2 dx=∫sin⁡(2x) dx2\int \frac{\sin{\left(2 x \right)}}{2}\, dx = \frac{\int \sin{\left(2 x \right)}\, dx}{2}

            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}

            So, the result is: −cos⁡(2x)4- \frac{\cos{\left(2 x \right)}}{4}

          Now evaluate the sub-integral.

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

          ∫(−cos⁡(2x)4) dx=−∫cos⁡(2x) dx4\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}

          1. Let u=2xu = 2 x.

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

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

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

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

          So, the result is: −sin⁡(2x)8- \frac{\sin{\left(2 x \right)}}{8}

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

          ∫(−πsin⁡(x)cos⁡(x)4) dx=−π∫sin⁡(x)cos⁡(x) dx4\int \left(- \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{4}\right)\, dx = - \frac{\pi \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx}{4}

          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}

          So, the result is: πcos⁡2(x)8\frac{\pi \cos^{2}{\left(x \right)}}{8}

        The result is: −xcos⁡(2x)4+sin⁡(2x)8+πcos⁡2(x)8- \frac{x \cos{\left(2 x \right)}}{4} + \frac{\sin{\left(2 x \right)}}{8} + \frac{\pi \cos^{2}{\left(x \right)}}{8}

      So, the result is: −xcos⁡(2x)2+sin⁡(2x)4+πcos⁡2(x)4- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos^{2}{\left(x \right)}}{4}

  2. Add the constant of integration:

    −xcos⁡(2x)2+sin⁡(2x)4+πcos⁡(2x)8+constant- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}+ \mathrm{constant}

The answer is:

−xcos⁡(2x)2+sin⁡(2x)4+πcos⁡(2x)8+constant- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                              
 |                                                               
 | /    pi\                   sin(2*x)   x*cos(2*x)   pi*cos(2*x)
 | |x - --|*sin(2*x) dx = C + -------- - ---------- + -----------
 | \    4 /                      4           2             8     
 |                                                               
/
∫(x−π4)sin⁡(2x) dx=C−xcos⁡(2x)2+sin⁡(2x)4+πcos⁡(2x)8\int \left(x - \frac{\pi}{4}\right) \sin{\left(2 x \right)}\, dx = C - \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}
Simplify
The graph
0.000.050.100.150.200.250.300.350.400.450.500.550.600.650.700.750.5-0.5
Plot the graph f(x)
The answer [src] 
1   pi
- - --
4   8
14−π8\frac{1}{4} - \frac{\pi}{8} 
=
= 
1   pi
- - --
4   8
14−π8\frac{1}{4} - \frac{\pi}{8} 
1/4 - pi/8
Numerical answer [src] 
-0.142699081698724
-0.142699081698724

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

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