Mister Exam

Other calculators

x*sinx*cos2x
Solving integrals/ x*sinx*cos2x

Integral of x*sinx*cos2x dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1                     
  /                     
 |                      
 |  x*sin(x)*cos(2*x) dx
 |                      
/                       
0
∫01xsin⁡(x)cos⁡(2x) dx\int\limits_{0}^{1} x \sin{\left(x \right)} \cos{\left(2 x \right)}\, dx 
Integral((x*sin(x))*cos(2*x), (x, 0, 1))
Detail solution

  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⁡(2x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(2 x \right)}.

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

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

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

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

          ∫(−u2) du\int \left(- u^{2}\right)\, du

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

            ∫u2 du=−∫u2 du\int u^{2}\, du = - \int u^{2}\, du

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

              ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            So, the result is: −u33- \frac{u^{3}}{3}

          Now substitute uu back in:

          −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

        So, the result is: −2cos⁡3(x)3- \frac{2 \cos^{3}{\left(x \right)}}{3}

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

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

        1. The integral of sine is negative cosine:

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

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

      The result is: −2cos⁡3(x)3+cos⁡(x)- \frac{2 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}

    Now evaluate the sub-integral.

  2. Integrate term-by-term:

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

      ∫(−2cos⁡3(x)3) dx=−2∫cos⁡3(x) dx3\int \left(- \frac{2 \cos^{3}{\left(x \right)}}{3}\right)\, dx = - \frac{2 \int \cos^{3}{\left(x \right)}\, dx}{3}

      1. Rewrite the integrand:

        cos⁡3(x)=(1−sin⁡2(x))cos⁡(x)\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}

      2. There are multiple ways to do this integral.

        Method #1

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

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

          ∫(1−u2) du\int \left(1 - u^{2}\right)\, du

          1. Integrate term-by-term:

            1. The integral of a constant is the constant times the variable of integration:

              ∫1 du=u\int 1\, du = u

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

              ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

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

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: −u33- \frac{u^{3}}{3}

            The result is: −u33+u- \frac{u^{3}}{3} + u

          Now substitute uu back in:

          −sin⁡3(x)3+sin⁡(x)- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

        Method #2

        1. Rewrite the integrand:

          (1−sin⁡2(x))cos⁡(x)=−sin⁡2(x)cos⁡(x)+cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} = - \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}

        2. Integrate term-by-term:

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

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

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

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

              ∫u2 du\int u^{2}\, du

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

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              Now substitute uu back in:

              sin⁡3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

            So, the result is: −sin⁡3(x)3- \frac{\sin^{3}{\left(x \right)}}{3}

          1. The integral of cosine is sine:

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

          The result is: −sin⁡3(x)3+sin⁡(x)- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

        Method #3

        1. Rewrite the integrand:

          (1−sin⁡2(x))cos⁡(x)=−sin⁡2(x)cos⁡(x)+cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} = - \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}

        2. Integrate term-by-term:

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

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

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

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

              ∫u2 du\int u^{2}\, du

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

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              Now substitute uu back in:

              sin⁡3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

            So, the result is: −sin⁡3(x)3- \frac{\sin^{3}{\left(x \right)}}{3}

          1. The integral of cosine is sine:

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

          The result is: −sin⁡3(x)3+sin⁡(x)- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

      So, the result is: 2sin⁡3(x)9−2sin⁡(x)3\frac{2 \sin^{3}{\left(x \right)}}{9} - \frac{2 \sin{\left(x \right)}}{3}

    1. The integral of cosine is sine:

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

    The result is: 2sin⁡3(x)9+sin⁡(x)3\frac{2 \sin^{3}{\left(x \right)}}{9} + \frac{\sin{\left(x \right)}}{3}

  3. Now simplify:

    −2xcos⁡3(x)3+xcos⁡(x)−2sin⁡3(x)9−sin⁡(x)3- \frac{2 x \cos^{3}{\left(x \right)}}{3} + x \cos{\left(x \right)} - \frac{2 \sin^{3}{\left(x \right)}}{9} - \frac{\sin{\left(x \right)}}{3}

  4. Add the constant of integration:

    −2xcos⁡3(x)3+xcos⁡(x)−2sin⁡3(x)9−sin⁡(x)3+constant- \frac{2 x \cos^{3}{\left(x \right)}}{3} + x \cos{\left(x \right)} - \frac{2 \sin^{3}{\left(x \right)}}{9} - \frac{\sin{\left(x \right)}}{3}+ \mathrm{constant}

The answer is:

−2xcos⁡3(x)3+xcos⁡(x)−2sin⁡3(x)9−sin⁡(x)3+constant- \frac{2 x \cos^{3}{\left(x \right)}}{3} + x \cos{\left(x \right)} - \frac{2 \sin^{3}{\left(x \right)}}{9} - \frac{\sin{\left(x \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                3                 /       3            \
 |                            2*sin (x)   sin(x)     |  2*cos (x)         |
 | x*sin(x)*cos(2*x) dx = C - --------- - ------ + x*|- --------- + cos(x)|
 |                                9         3        \      3             /
/
∫xsin⁡(x)cos⁡(2x) dx=C+x(−2cos⁡3(x)3+cos⁡(x))−2sin⁡3(x)9−sin⁡(x)3\int x \sin{\left(x \right)} \cos{\left(2 x \right)}\, dx = C + x \left(- \frac{2 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}\right) - \frac{2 \sin^{3}{\left(x \right)}}{9} - \frac{\sin{\left(x \right)}}{3}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.5-0.5
Plot the graph f(x)
The answer [src] 
  4*cos(1)*sin(2)   cos(1)*cos(2)   2*sin(1)*sin(2)   5*cos(2)*sin(1)
- --------------- + ------------- + --------------- + ---------------
         9                3                3                 9
−4sin⁡(2)cos⁡(1)9+5sin⁡(1)cos⁡(2)9+cos⁡(1)cos⁡(2)3+2sin⁡(1)sin⁡(2)3- \frac{4 \sin{\left(2 \right)} \cos{\left(1 \right)}}{9} + \frac{5 \sin{\left(1 \right)} \cos{\left(2 \right)}}{9} + \frac{\cos{\left(1 \right)} \cos{\left(2 \right)}}{3} + \frac{2 \sin{\left(1 \right)} \sin{\left(2 \right)}}{3} 
=
= 
  4*cos(1)*sin(2)   cos(1)*cos(2)   2*sin(1)*sin(2)   5*cos(2)*sin(1)
- --------------- + ------------- + --------------- + ---------------
         9                3                3                 9
−4sin⁡(2)cos⁡(1)9+5sin⁡(1)cos⁡(2)9+cos⁡(1)cos⁡(2)3+2sin⁡(1)sin⁡(2)3- \frac{4 \sin{\left(2 \right)} \cos{\left(1 \right)}}{9} + \frac{5 \sin{\left(1 \right)} \cos{\left(2 \right)}}{9} + \frac{\cos{\left(1 \right)} \cos{\left(2 \right)}}{3} + \frac{2 \sin{\left(1 \right)} \sin{\left(2 \right)}}{3} 
-4*cos(1)*sin(2)/9 + cos(1)*cos(2)/3 + 2*sin(1)*sin(2)/3 + 5*cos(2)*sin(1)/9
Numerical answer [src] 
0.0222544104112996
0.0222544104112996
The graph
Integral of x*sinx*cos2x dx

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

    © Mister Exam – Calculator