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x^2sinx^3dx
Solving integrals/ x^2sinx^3dx

Integral of x^2sinx^3dx dx

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∫01x2sin⁡3(x)1 dx\int\limits_{0}^{1} x^{2} \sin^{3}{\left(x \right)} 1\, dx 
Integral(x^2*sin(x)^3*1, (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)=x2u{\left(x \right)} = x^{2} and let dv⁡(x)=sin⁡3(x)\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)}.

    Then du⁡(x)=2x\operatorname{du}{\left(x \right)} = 2 x.

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

    1. Rewrite the integrand:

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

    2. 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 dudu:

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

        1. Integrate term-by-term:

          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}

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

            ∫(−1) du=−u\int \left(-1\right)\, du = - u

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

        Now substitute uu back in:

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

      Method #2

      1. Rewrite the integrand:

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

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

            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}

            Now substitute uu back in:

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

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

        1. The integral of sine is negative cosine:

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

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

      Method #3

      1. Rewrite the integrand:

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

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

            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}

            Now substitute uu back in:

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

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

        1. The integral of sine is negative cosine:

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

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

    Now evaluate the sub-integral.

  2. 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)=2x3u{\left(x \right)} = \frac{2 x}{3} and let dv⁡(x)=cos⁡3(x)−3cos⁡(x)\operatorname{dv}{\left(x \right)} = \cos^{3}{\left(x \right)} - 3 \cos{\left(x \right)}.

    Then du⁡(x)=23\operatorname{du}{\left(x \right)} = \frac{2}{3}.

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

    1. Integrate term-by-term:

      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. 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)}

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

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

        1. The integral of cosine is sine:

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

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

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

    Now evaluate the sub-integral.

  3. Integrate term-by-term:

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

      ∫(−2sin⁡3(x)9) dx=−2∫sin⁡3(x) dx9\int \left(- \frac{2 \sin^{3}{\left(x \right)}}{9}\right)\, dx = - \frac{2 \int \sin^{3}{\left(x \right)}\, dx}{9}

      1. Rewrite the integrand:

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

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

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

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

        1. Integrate term-by-term:

          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}

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

            ∫(−1) du=−u\int \left(-1\right)\, du = - u

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

        Now substitute uu back in:

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

      So, the result is: −2cos⁡3(x)27+2cos⁡(x)9- \frac{2 \cos^{3}{\left(x \right)}}{27} + \frac{2 \cos{\left(x \right)}}{9}

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

      ∫(−4sin⁡(x)3) dx=−4∫sin⁡(x) dx3\int \left(- \frac{4 \sin{\left(x \right)}}{3}\right)\, dx = - \frac{4 \int \sin{\left(x \right)}\, dx}{3}

      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: 4cos⁡(x)3\frac{4 \cos{\left(x \right)}}{3}

    The result is: −2cos⁡3(x)27+14cos⁡(x)9- \frac{2 \cos^{3}{\left(x \right)}}{27} + \frac{14 \cos{\left(x \right)}}{9}

  4. Now simplify:

    x2(cos⁡2(x)−3)cos⁡(x)3+2x(sin⁡2(x)+6)sin⁡(x)9−2cos⁡3(x)27+14cos⁡(x)9\frac{x^{2} \left(\cos^{2}{\left(x \right)} - 3\right) \cos{\left(x \right)}}{3} + \frac{2 x \left(\sin^{2}{\left(x \right)} + 6\right) \sin{\left(x \right)}}{9} - \frac{2 \cos^{3}{\left(x \right)}}{27} + \frac{14 \cos{\left(x \right)}}{9}

  5. Add the constant of integration:

    x2(cos⁡2(x)−3)cos⁡(x)3+2x(sin⁡2(x)+6)sin⁡(x)9−2cos⁡3(x)27+14cos⁡(x)9+constant\frac{x^{2} \left(\cos^{2}{\left(x \right)} - 3\right) \cos{\left(x \right)}}{3} + \frac{2 x \left(\sin^{2}{\left(x \right)} + 6\right) \sin{\left(x \right)}}{9} - \frac{2 \cos^{3}{\left(x \right)}}{27} + \frac{14 \cos{\left(x \right)}}{9}+ \mathrm{constant}

The answer is:

x2(cos⁡2(x)−3)cos⁡(x)3+2x(sin⁡2(x)+6)sin⁡(x)9−2cos⁡3(x)27+14cos⁡(x)9+constant\frac{x^{2} \left(\cos^{2}{\left(x \right)} - 3\right) \cos{\left(x \right)}}{3} + \frac{2 x \left(\sin^{2}{\left(x \right)} + 6\right) \sin{\left(x \right)}}{9} - \frac{2 \cos^{3}{\left(x \right)}}{27} + \frac{14 \cos{\left(x \right)}}{9}+ \mathrm{constant}

The answer (Indefinite) [src] 
                                                                              /               3   \
  /                                                                           |            sin (x)|
 |                            3                     /             3   \   2*x*|-2*sin(x) - -------|
 |  2    3               2*cos (x)   14*cos(x)    2 |          cos (x)|       \               3   /
 | x *sin (x)*1 dx = C - --------- + --------- + x *|-cos(x) + -------| - -------------------------
 |                           27          9          \             3   /               3            
/
−6 x sin⁡(3 x)+(2−9 x2) cos⁡(3 x)−162 x sin⁡x+(81 x2−162) cos⁡x108-{{6\,x\,\sin \left(3\,x\right)+\left(2-9\,x^2\right)\,\cos \left(3 \,x\right)-162\,x\,\sin x+\left(81\,x^2-162\right)\,\cos x}\over{108 }}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9002
Plot the graph f(x)
The answer [src] 
             3            3           2                  2          
  40   14*sin (1)   22*cos (1)   4*cos (1)*sin(1)   5*sin (1)*cos(1)
- -- + ---------- + ---------- + ---------------- + ----------------
  27       9            27              3                  9
−6 sin⁡3−7 cos⁡3−162 sin⁡1−81 cos⁡1108−4027-{{6\,\sin 3-7\,\cos 3-162\,\sin 1-81\,\cos 1}\over{108}}-{{40 }\over{27}} 
=
= 
             3            3           2                  2          
  40   14*sin (1)   22*cos (1)   4*cos (1)*sin(1)   5*sin (1)*cos(1)
- -- + ---------- + ---------- + ---------------- + ----------------
  27       9            27              3                  9
−4027+22cos⁡3(1)27+5sin⁡2(1)cos⁡(1)9+4sin⁡(1)cos⁡2(1)3+14sin⁡3(1)9- \frac{40}{27} + \frac{22 \cos^{3}{\left(1 \right)}}{27} + \frac{5 \sin^{2}{\left(1 \right)} \cos{\left(1 \right)}}{9} + \frac{4 \sin{\left(1 \right)} \cos^{2}{\left(1 \right)}}{3} + \frac{14 \sin^{3}{\left(1 \right)}}{9}
Simplify
Numerical answer [src] 
0.113945544348484
0.113945544348484
The graph
Integral of x^2sinx^3dx dx

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

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