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sin^4(x)cos^2(x)
Solving integrals/ sin^4(x)cos^2(x)

Integral of sin^4(x)cos^2(x) dx

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∫01sin⁡4(x)cos⁡2(x) dx\int\limits_{0}^{1} \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx 
Integral(sin(x)^4*cos(x)^2, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

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

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=2xu = 2 x.

      Then let du=2dxdu = 2 dx and substitute dudu:

      ∫(cos⁡3(u)16−cos⁡2(u)16−cos⁡(u)16+116) du\int \left(\frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\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:

          ∫cos⁡3(u)16 du=∫cos⁡3(u) du16\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}

          1. Rewrite the integrand:

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

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

            Then let du=cos⁡(u)dudu = \cos{\left(u \right)} du 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(u)3+sin⁡(u)- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}

          So, the result is: −sin⁡3(u)48+sin⁡(u)16- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}

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

          ∫(−cos⁡2(u)16) du=−∫cos⁡2(u) du16\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}

          1. Rewrite the integrand:

            cos⁡2(u)=cos⁡(2u)2+12\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \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⁡(2u)2 du=∫cos⁡(2u) du2\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}

              1. Let u=2uu = 2 u.

                Then let du=2dudu = 2 du 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⁡(2u)2\frac{\sin{\left(2 u \right)}}{2}

              So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{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: u2+sin⁡(2u)4\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}

          So, the result is: −u32−sin⁡(2u)64- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}

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

          ∫(−cos⁡(u)16) du=−∫cos⁡(u) du16\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}

          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)16- \frac{\sin{\left(u \right)}}{16}

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

          ∫116 du=u16\int \frac{1}{16}\, du = \frac{u}{16}

        The result is: u32−sin⁡(2u)64−sin⁡3(u)48\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} - \frac{\sin^{3}{\left(u \right)}}{48}

      Now substitute uu back in:

      x16−sin⁡3(2x)48−sin⁡(4x)64\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}

    Method #2

    1. Rewrite the integrand:

      (12−cos⁡(2x)2)2(cos⁡(2x)2+12)=cos⁡3(2x)8−cos⁡2(2x)8−cos⁡(2x)8+18\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(2 x \right)}}{8} - \frac{\cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{8} + \frac{1}{8}

    2. Integrate term-by-term:

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

        ∫cos⁡3(2x)8 dx=∫cos⁡3(2x) dx8\int \frac{\cos^{3}{\left(2 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}

        1. Rewrite the integrand:

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

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

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

          ∫(12−u22) du\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du

          1. Integrate term-by-term:

            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}

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

              ∫(−u22) du=−∫u2 du2\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}

              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: −u36- \frac{u^{3}}{6}

            The result is: −u36+u2- \frac{u^{3}}{6} + \frac{u}{2}

          Now substitute uu back in:

          −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡3(2x)48+sin⁡(2x)16- \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{16}

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

        ∫(−cos⁡2(2x)8) dx=−∫cos⁡2(2x) dx8\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{8}

        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: −x16−sin⁡(4x)64- \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64}

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

        ∫(−cos⁡(2x)8) dx=−∫cos⁡(2x) dx8\int \left(- \frac{\cos{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{8}

        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)16- \frac{\sin{\left(2 x \right)}}{16}

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

        ∫18 dx=x8\int \frac{1}{8}\, dx = \frac{x}{8}

      The result is: x16−sin⁡3(2x)48−sin⁡(4x)64\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}

    Method #3

    1. Rewrite the integrand:

      (12−cos⁡(2x)2)2(cos⁡(2x)2+12)=cos⁡3(2x)8−cos⁡2(2x)8−cos⁡(2x)8+18\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(2 x \right)}}{8} - \frac{\cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{8} + \frac{1}{8}

    2. Integrate term-by-term:

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

        ∫cos⁡3(2x)8 dx=∫cos⁡3(2x) dx8\int \frac{\cos^{3}{\left(2 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}

        1. Rewrite the integrand:

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

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

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

          ∫(12−u22) du\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du

          1. Integrate term-by-term:

            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}

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

              ∫(−u22) du=−∫u2 du2\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}

              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: −u36- \frac{u^{3}}{6}

            The result is: −u36+u2- \frac{u^{3}}{6} + \frac{u}{2}

          Now substitute uu back in:

          −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡3(2x)48+sin⁡(2x)16- \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{16}

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

        ∫(−cos⁡2(2x)8) dx=−∫cos⁡2(2x) dx8\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{8}

        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: −x16−sin⁡(4x)64- \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64}

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

        ∫(−cos⁡(2x)8) dx=−∫cos⁡(2x) dx8\int \left(- \frac{\cos{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{8}

        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)16- \frac{\sin{\left(2 x \right)}}{16}

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

        ∫18 dx=x8\int \frac{1}{8}\, dx = \frac{x}{8}

      The result is: x16−sin⁡3(2x)48−sin⁡(4x)64\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}

  3. Add the constant of integration:

    x16−sin⁡3(2x)48−sin⁡(4x)64+constant\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}+ \mathrm{constant}

The answer is:

x16−sin⁡3(2x)48−sin⁡(4x)64+constant\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                  
 |                             3                     
 |    4       2             sin (2*x)   sin(4*x)   x 
 | sin (x)*cos (x) dx = C - --------- - -------- + --
 |                              48         64      16
/
2 x−sin⁡(4 x)24−sin⁡3(2 x)68{{{{2\,x-{{\sin \left(4\,x\right)}\over{2}}}\over{4}}-{{\sin ^3 \left(2\,x\right)}\over{6}}}\over{8}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.2
Plot the graph f(x)
The answer [src] 
                        3                5          
1    cos(1)*sin(1)   sin (1)*cos(1)   sin (1)*cos(1)
-- - ------------- - -------------- + --------------
16         16              24               6
−3 sin⁡4+4 sin⁡32−12192-{{3\,\sin 4+4\,\sin ^32-12}\over{192}} 
=
= 
                        3                5          
1    cos(1)*sin(1)   sin (1)*cos(1)   sin (1)*cos(1)
-- - ------------- - -------------- + --------------
16         16              24               6
−sin⁡(1)cos⁡(1)16−sin⁡3(1)cos⁡(1)24+sin⁡5(1)cos⁡(1)6+116- \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{16} - \frac{\sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{24} + \frac{\sin^{5}{\left(1 \right)} \cos{\left(1 \right)}}{6} + \frac{1}{16}
Simplify
Numerical answer [src] 
0.0586619776419157
0.0586619776419157
The graph
Integral of sin^4(x)cos^2(x) dx

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

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