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cos^2(3x)sin^4(3x)

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Solving integrals/ cos^2(3x)sin^4(3x)

Integral of cos^2(3x)sin^4(3x) dx

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  0                       
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 |  cos (3*x)*sin (3*x) dx
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0
∫00sin⁡4(3x)cos⁡2(3x) dx\int\limits_{0}^{0} \sin^{4}{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\, dx 
Integral(cos(3*x)^2*sin(3*x)^4, (x, 0, 0))
Detail solution

  1. Rewrite the integrand:

    sin⁡4(3x)cos⁡2(3x)=(12−cos⁡(6x)2)2(cos⁡(6x)2+12)\sin^{4}{\left(3 x \right)} \cos^{2}{\left(3 x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(6 x \right)}}{2} + \frac{1}{2}\right)

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=6xu = 6 x.

      Then let du=6dxdu = 6 dx and substitute dudu:

      ∫(cos⁡3(u)48−cos⁡2(u)48−cos⁡(u)48+148) du\int \left(\frac{\cos^{3}{\left(u \right)}}{48} - \frac{\cos^{2}{\left(u \right)}}{48} - \frac{\cos{\left(u \right)}}{48} + \frac{1}{48}\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)48 du=∫cos⁡3(u) du48\int \frac{\cos^{3}{\left(u \right)}}{48}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{48}

          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)144+sin⁡(u)48- \frac{\sin^{3}{\left(u \right)}}{144} + \frac{\sin{\left(u \right)}}{48}

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

          ∫(−cos⁡2(u)48) du=−∫cos⁡2(u) du48\int \left(- \frac{\cos^{2}{\left(u \right)}}{48}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{48}

          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)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⁡(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: −u96−sin⁡(2u)192- \frac{u}{96} - \frac{\sin{\left(2 u \right)}}{192}

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

          ∫(−cos⁡(u)48) du=−∫cos⁡(u) du48\int \left(- \frac{\cos{\left(u \right)}}{48}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{48}

          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)48- \frac{\sin{\left(u \right)}}{48}

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

          ∫148 du=u48\int \frac{1}{48}\, du = \frac{u}{48}

        The result is: u96−sin⁡(2u)192−sin⁡3(u)144\frac{u}{96} - \frac{\sin{\left(2 u \right)}}{192} - \frac{\sin^{3}{\left(u \right)}}{144}

      Now substitute uu back in:

      x16−sin⁡3(6x)144−sin⁡(12x)192\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}

    Method #2

    1. Rewrite the integrand:

      (12−cos⁡(6x)2)2(cos⁡(6x)2+12)=cos⁡3(6x)8−cos⁡2(6x)8−cos⁡(6x)8+18\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(6 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(6 x \right)}}{8} - \frac{\cos^{2}{\left(6 x \right)}}{8} - \frac{\cos{\left(6 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(6x)8 dx=∫cos⁡3(6x) dx8\int \frac{\cos^{3}{\left(6 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(6 x \right)}\, dx}{8}

        1. Rewrite the integrand:

          cos⁡3(6x)=(1−sin⁡2(6x))cos⁡(6x)\cos^{3}{\left(6 x \right)} = \left(1 - \sin^{2}{\left(6 x \right)}\right) \cos{\left(6 x \right)}

        2. Let u=sin⁡(6x)u = \sin{\left(6 x \right)}.

          Then let du=6cos⁡(6x)dxdu = 6 \cos{\left(6 x \right)} dx and substitute dudu:

          ∫(16−u26) du\int \left(\frac{1}{6} - \frac{u^{2}}{6}\right)\, du

          1. Integrate term-by-term:

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

              ∫16 du=u6\int \frac{1}{6}\, du = \frac{u}{6}

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

              ∫(−u26) du=−∫u2 du6\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}

              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: −u318- \frac{u^{3}}{18}

            The result is: −u318+u6- \frac{u^{3}}{18} + \frac{u}{6}

          Now substitute uu back in:

          −sin⁡3(6x)18+sin⁡(6x)6- \frac{\sin^{3}{\left(6 x \right)}}{18} + \frac{\sin{\left(6 x \right)}}{6}

        So, the result is: −sin⁡3(6x)144+sin⁡(6x)48- \frac{\sin^{3}{\left(6 x \right)}}{144} + \frac{\sin{\left(6 x \right)}}{48}

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

        ∫(−cos⁡2(6x)8) dx=−∫cos⁡2(6x) dx8\int \left(- \frac{\cos^{2}{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{8}

        1. Rewrite the integrand:

          cos⁡2(6x)=cos⁡(12x)2+12\cos^{2}{\left(6 x \right)} = \frac{\cos{\left(12 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⁡(12x)2 dx=∫cos⁡(12x) dx2\int \frac{\cos{\left(12 x \right)}}{2}\, dx = \frac{\int \cos{\left(12 x \right)}\, dx}{2}

            1. Let u=12xu = 12 x.

              Then let du=12dxdu = 12 dx and substitute du12\frac{du}{12}:

              ∫cos⁡(u)12 du\int \frac{\cos{\left(u \right)}}{12}\, du

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

                ∫cos⁡(u) du=∫cos⁡(u) du12\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}

                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)12\frac{\sin{\left(u \right)}}{12}

              Now substitute uu back in:

              sin⁡(12x)12\frac{\sin{\left(12 x \right)}}{12}

            So, the result is: sin⁡(12x)24\frac{\sin{\left(12 x \right)}}{24}

          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⁡(12x)24\frac{x}{2} + \frac{\sin{\left(12 x \right)}}{24}

        So, the result is: −x16−sin⁡(12x)192- \frac{x}{16} - \frac{\sin{\left(12 x \right)}}{192}

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

        ∫(−cos⁡(6x)8) dx=−∫cos⁡(6x) dx8\int \left(- \frac{\cos{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{8}

        1. Let u=6xu = 6 x.

          Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

          ∫cos⁡(u)6 du\int \frac{\cos{\left(u \right)}}{6}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du6\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

            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)6\frac{\sin{\left(u \right)}}{6}

          Now substitute uu back in:

          sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

        So, the result is: −sin⁡(6x)48- \frac{\sin{\left(6 x \right)}}{48}

      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(6x)144−sin⁡(12x)192\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}

    Method #3

    1. Rewrite the integrand:

      (12−cos⁡(6x)2)2(cos⁡(6x)2+12)=cos⁡3(6x)8−cos⁡2(6x)8−cos⁡(6x)8+18\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(6 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(6 x \right)}}{8} - \frac{\cos^{2}{\left(6 x \right)}}{8} - \frac{\cos{\left(6 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(6x)8 dx=∫cos⁡3(6x) dx8\int \frac{\cos^{3}{\left(6 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(6 x \right)}\, dx}{8}

        1. Rewrite the integrand:

          cos⁡3(6x)=(1−sin⁡2(6x))cos⁡(6x)\cos^{3}{\left(6 x \right)} = \left(1 - \sin^{2}{\left(6 x \right)}\right) \cos{\left(6 x \right)}

        2. Let u=sin⁡(6x)u = \sin{\left(6 x \right)}.

          Then let du=6cos⁡(6x)dxdu = 6 \cos{\left(6 x \right)} dx and substitute dudu:

          ∫(16−u26) du\int \left(\frac{1}{6} - \frac{u^{2}}{6}\right)\, du

          1. Integrate term-by-term:

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

              ∫16 du=u6\int \frac{1}{6}\, du = \frac{u}{6}

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

              ∫(−u26) du=−∫u2 du6\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}

              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: −u318- \frac{u^{3}}{18}

            The result is: −u318+u6- \frac{u^{3}}{18} + \frac{u}{6}

          Now substitute uu back in:

          −sin⁡3(6x)18+sin⁡(6x)6- \frac{\sin^{3}{\left(6 x \right)}}{18} + \frac{\sin{\left(6 x \right)}}{6}

        So, the result is: −sin⁡3(6x)144+sin⁡(6x)48- \frac{\sin^{3}{\left(6 x \right)}}{144} + \frac{\sin{\left(6 x \right)}}{48}

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

        ∫(−cos⁡2(6x)8) dx=−∫cos⁡2(6x) dx8\int \left(- \frac{\cos^{2}{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{8}

        1. Rewrite the integrand:

          cos⁡2(6x)=cos⁡(12x)2+12\cos^{2}{\left(6 x \right)} = \frac{\cos{\left(12 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⁡(12x)2 dx=∫cos⁡(12x) dx2\int \frac{\cos{\left(12 x \right)}}{2}\, dx = \frac{\int \cos{\left(12 x \right)}\, dx}{2}

            1. Let u=12xu = 12 x.

              Then let du=12dxdu = 12 dx and substitute du12\frac{du}{12}:

              ∫cos⁡(u)12 du\int \frac{\cos{\left(u \right)}}{12}\, du

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

                ∫cos⁡(u) du=∫cos⁡(u) du12\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}

                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)12\frac{\sin{\left(u \right)}}{12}

              Now substitute uu back in:

              sin⁡(12x)12\frac{\sin{\left(12 x \right)}}{12}

            So, the result is: sin⁡(12x)24\frac{\sin{\left(12 x \right)}}{24}

          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⁡(12x)24\frac{x}{2} + \frac{\sin{\left(12 x \right)}}{24}

        So, the result is: −x16−sin⁡(12x)192- \frac{x}{16} - \frac{\sin{\left(12 x \right)}}{192}

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

        ∫(−cos⁡(6x)8) dx=−∫cos⁡(6x) dx8\int \left(- \frac{\cos{\left(6 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{8}

        1. Let u=6xu = 6 x.

          Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

          ∫cos⁡(u)6 du\int \frac{\cos{\left(u \right)}}{6}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du6\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

            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)6\frac{\sin{\left(u \right)}}{6}

          Now substitute uu back in:

          sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

        So, the result is: −sin⁡(6x)48- \frac{\sin{\left(6 x \right)}}{48}

      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(6x)144−sin⁡(12x)192\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}

  3. Add the constant of integration:

    x16−sin⁡3(6x)144−sin⁡(12x)192+constant\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}+ \mathrm{constant}

The answer is:

x16−sin⁡3(6x)144−sin⁡(12x)192+constant\frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                       
 |                                 3                      
 |    2         4               sin (6*x)   sin(12*x)   x 
 | cos (3*x)*sin (3*x) dx = C - --------- - --------- + --
 |                                 144         192      16
/
∫sin⁡4(3x)cos⁡2(3x) dx=C+x16−sin⁡3(6x)144−sin⁡(12x)192\int \sin^{4}{\left(3 x \right)} \cos^{2}{\left(3 x \right)}\, dx = C + \frac{x}{16} - \frac{\sin^{3}{\left(6 x \right)}}{144} - \frac{\sin{\left(12 x \right)}}{192}
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Integral of cos^2(3x)sin^4(3x) dx

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

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