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(sin^3)xcosxdx
Solving integrals/ (sin^3)xcosxdx

Integral of (sin^3)xcosxdx dx

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

        ∫u3 du\int u^{3}\, du

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

          ∫u3 du=u44\int u^{3}\, du = \frac{u^{4}}{4}

        Now substitute uu back in:

        sin⁡4(x)4\frac{\sin^{4}{\left(x \right)}}{4}

      Method #2

      1. Rewrite the integrand:

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

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

        Then let du=2sin⁡(x)cos⁡(x)dxdu = 2 \sin{\left(x \right)} \cos{\left(x \right)} dx and substitute dudu:

        ∫(u2+12) du\int \left(\frac{u}{2} + \frac{1}{2}\right)\, du

        1. Integrate term-by-term:

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

            ∫u2 du=∫u du2\int \frac{u}{2}\, du = \frac{\int u\, du}{2}

            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: u24\frac{u^{2}}{4}

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

            ∫12 du=u2\int \frac{1}{2}\, du = \frac{u}{2}

          The result is: u24+u2\frac{u^{2}}{4} + \frac{u}{2}

        Now substitute uu back in:

        cos⁡4(x)4−cos⁡2(x)2\frac{\cos^{4}{\left(x \right)}}{4} - \frac{\cos^{2}{\left(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:

    ∫sin⁡4(x)4 dx=∫sin⁡4(x) dx4\int \frac{\sin^{4}{\left(x \right)}}{4}\, dx = \frac{\int \sin^{4}{\left(x \right)}\, dx}{4}

    1. Rewrite the integrand:

      sin⁡4(x)=(12−cos⁡(2x)2)2\sin^{4}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2}

    2. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

        (12−cos⁡(2x)2)2=cos⁡2(2x)4−cos⁡(2x)2+14\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}

      2. Integrate term-by-term:

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

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

          1. Rewrite the integrand:

            cos⁡2(2x)=cos⁡(4x)2+12\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}

          2. Integrate term-by-term:

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

              ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

              1. Let u=4xu = 4 x.

                Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

                ∫cos⁡(u)16 du\int \frac{\cos{\left(u \right)}}{16}\, du

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

                  ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                  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)4\frac{\sin{\left(u \right)}}{4}

                Now substitute uu back in:

                sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

              So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

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

              ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

            The result is: x2+sin⁡(4x)8\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}

          So, the result is: x8+sin⁡(4x)32\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}

        1. 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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

              ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, 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 is the constant times the variable of integration:

          ∫14 dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

        The result is: 3x8−sin⁡(2x)4+sin⁡(4x)32\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}

      Method #2

      1. Rewrite the integrand:

        (12−cos⁡(2x)2)2=cos⁡2(2x)4−cos⁡(2x)2+14\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}

      2. Integrate term-by-term:

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

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

          1. Rewrite the integrand:

            cos⁡2(2x)=cos⁡(4x)2+12\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}

          2. Integrate term-by-term:

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

              ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

              1. Let u=4xu = 4 x.

                Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

                ∫cos⁡(u)16 du\int \frac{\cos{\left(u \right)}}{16}\, du

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

                  ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                  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)4\frac{\sin{\left(u \right)}}{4}

                Now substitute uu back in:

                sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

              So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

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

              ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

            The result is: x2+sin⁡(4x)8\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}

          So, the result is: x8+sin⁡(4x)32\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}

        1. 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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

              ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, 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 is the constant times the variable of integration:

          ∫14 dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

        The result is: 3x8−sin⁡(2x)4+sin⁡(4x)32\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}

    So, the result is: 3x32−sin⁡(2x)16+sin⁡(4x)128\frac{3 x}{32} - \frac{\sin{\left(2 x \right)}}{16} + \frac{\sin{\left(4 x \right)}}{128}

  3. Add the constant of integration:

    xsin⁡4(x)4−3x32+sin⁡(2x)16−sin⁡(4x)128+constant\frac{x \sin^{4}{\left(x \right)}}{4} - \frac{3 x}{32} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{128}+ \mathrm{constant}

The answer is:

xsin⁡4(x)4−3x32+sin⁡(2x)16−sin⁡(4x)128+constant\frac{x \sin^{4}{\left(x \right)}}{4} - \frac{3 x}{32} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{128}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                 
 |                                                              4   
 |    3                        3*x   sin(4*x)   sin(2*x)   x*sin (x)
 | sin (x)*x*cos(x)*1 dx = C - --- - -------- + -------- + ---------
 |                              32     128         16          4    
/
−sin⁡(4 x)−4 x cos⁡(4 x)−8 sin⁡(2 x)+16 x cos⁡(2 x)128-{{\sin \left(4\,x\right)-4\,x\,\cos \left(4\,x\right)-8\,\sin \left(2\,x\right)+16\,x\,\cos \left(2\,x\right)}\over{128}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.5
Plot the graph f(x)
The answer [src] 
       4           4           2       2           3                  3          
  3*cos (1)   5*sin (1)   3*cos (1)*sin (1)   3*cos (1)*sin(1)   5*sin (1)*cos(1)
- --------- + --------- - ----------------- + ---------------- + ----------------
      32          32              16                 32                 32
−sin⁡4−4 cos⁡4−8 sin⁡2+16 cos⁡2128-{{\sin 4-4\,\cos 4-8\,\sin 2+16\,\cos 2}\over{128}} 
=
= 
       4           4           2       2           3                  3          
  3*cos (1)   5*sin (1)   3*cos (1)*sin (1)   3*cos (1)*sin(1)   5*sin (1)*cos(1)
- --------- + --------- - ----------------- + ---------------- + ----------------
      32          32              16                 32                 32
−3sin⁡2(1)cos⁡2(1)16−3cos⁡4(1)32+3sin⁡(1)cos⁡3(1)32+5sin⁡3(1)cos⁡(1)32+5sin⁡4(1)32- \frac{3 \sin^{2}{\left(1 \right)} \cos^{2}{\left(1 \right)}}{16} - \frac{3 \cos^{4}{\left(1 \right)}}{32} + \frac{3 \sin{\left(1 \right)} \cos^{3}{\left(1 \right)}}{32} + \frac{5 \sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{32} + \frac{5 \sin^{4}{\left(1 \right)}}{32}
Simplify
Numerical answer [src] 
0.0943356000876032
0.0943356000876032
The graph
Integral of (sin^3)xcosxdx dx

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

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