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Solving integrals/ cos(5x)*(cos(2x))^2

Integral of cos(5x)*(cos(2x))^2 dx

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∫11cos⁡2(2x)cos⁡(5x) dx\int\limits_{1}^{1} \cos^{2}{\left(2 x \right)} \cos{\left(5 x \right)}\, dx 
Integral(cos(5*x)*cos(2*x)^2, (x, 1, 1))
Detail solution

  1. Rewrite the integrand:

    cos⁡2(2x)cos⁡(5x)=64cos⁡9(x)−144cos⁡7(x)+116cos⁡5(x)−40cos⁡3(x)+5cos⁡(x)\cos^{2}{\left(2 x \right)} \cos{\left(5 x \right)} = 64 \cos^{9}{\left(x \right)} - 144 \cos^{7}{\left(x \right)} + 116 \cos^{5}{\left(x \right)} - 40 \cos^{3}{\left(x \right)} + 5 \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:

      ∫64cos⁡9(x) dx=64∫cos⁡9(x) dx\int 64 \cos^{9}{\left(x \right)}\, dx = 64 \int \cos^{9}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        cos⁡9(x)=(1−sin⁡2(x))4cos⁡(x)\cos^{9}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)}

      2. There are multiple ways to do this integral.

        Method #1

        1. Rewrite the integrand:

          (1−sin⁡2(x))4cos⁡(x)=sin⁡8(x)cos⁡(x)−4sin⁡6(x)cos⁡(x)+6sin⁡4(x)cos⁡(x)−4sin⁡2(x)cos⁡(x)+cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}

        2. Integrate term-by-term:

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

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

            ∫u8 du\int u^{8}\, du

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

              ∫u8 du=u99\int u^{8}\, du = \frac{u^{9}}{9}

            Now substitute uu back in:

            sin⁡9(x)9\frac{\sin^{9}{\left(x \right)}}{9}

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

            ∫(−4sin⁡6(x)cos⁡(x)) dx=−4∫sin⁡6(x)cos⁡(x) dx\int \left(- 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{6}{\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:

              ∫u6 du\int u^{6}\, du

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

                ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

              Now substitute uu back in:

              sin⁡7(x)7\frac{\sin^{7}{\left(x \right)}}{7}

            So, the result is: −4sin⁡7(x)7- \frac{4 \sin^{7}{\left(x \right)}}{7}

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

            ∫6sin⁡4(x)cos⁡(x) dx=6∫sin⁡4(x)cos⁡(x) dx\int 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{4}{\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:

              ∫u4 du\int u^{4}\, du

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

                ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

              Now substitute uu back in:

              sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

            So, the result is: 6sin⁡5(x)5\frac{6 \sin^{5}{\left(x \right)}}{5}

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

            ∫(−4sin⁡2(x)cos⁡(x)) dx=−4∫sin⁡2(x)cos⁡(x) dx\int \left(- 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \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: −4sin⁡3(x)3- \frac{4 \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⁡9(x)9−4sin⁡7(x)7+6sin⁡5(x)5−4sin⁡3(x)3+sin⁡(x)\frac{\sin^{9}{\left(x \right)}}{9} - \frac{4 \sin^{7}{\left(x \right)}}{7} + \frac{6 \sin^{5}{\left(x \right)}}{5} - \frac{4 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

        Method #2

        1. Rewrite the integrand:

          (1−sin⁡2(x))4cos⁡(x)=sin⁡8(x)cos⁡(x)−4sin⁡6(x)cos⁡(x)+6sin⁡4(x)cos⁡(x)−4sin⁡2(x)cos⁡(x)+cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}

        2. Integrate term-by-term:

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

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

            ∫u8 du\int u^{8}\, du

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

              ∫u8 du=u99\int u^{8}\, du = \frac{u^{9}}{9}

            Now substitute uu back in:

            sin⁡9(x)9\frac{\sin^{9}{\left(x \right)}}{9}

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

            ∫(−4sin⁡6(x)cos⁡(x)) dx=−4∫sin⁡6(x)cos⁡(x) dx\int \left(- 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{6}{\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:

              ∫u6 du\int u^{6}\, du

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

                ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

              Now substitute uu back in:

              sin⁡7(x)7\frac{\sin^{7}{\left(x \right)}}{7}

            So, the result is: −4sin⁡7(x)7- \frac{4 \sin^{7}{\left(x \right)}}{7}

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

            ∫6sin⁡4(x)cos⁡(x) dx=6∫sin⁡4(x)cos⁡(x) dx\int 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{4}{\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:

              ∫u4 du\int u^{4}\, du

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

                ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

              Now substitute uu back in:

              sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

            So, the result is: 6sin⁡5(x)5\frac{6 \sin^{5}{\left(x \right)}}{5}

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

            ∫(−4sin⁡2(x)cos⁡(x)) dx=−4∫sin⁡2(x)cos⁡(x) dx\int \left(- 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \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: −4sin⁡3(x)3- \frac{4 \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⁡9(x)9−4sin⁡7(x)7+6sin⁡5(x)5−4sin⁡3(x)3+sin⁡(x)\frac{\sin^{9}{\left(x \right)}}{9} - \frac{4 \sin^{7}{\left(x \right)}}{7} + \frac{6 \sin^{5}{\left(x \right)}}{5} - \frac{4 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

      So, the result is: 64sin⁡9(x)9−256sin⁡7(x)7+384sin⁡5(x)5−256sin⁡3(x)3+64sin⁡(x)\frac{64 \sin^{9}{\left(x \right)}}{9} - \frac{256 \sin^{7}{\left(x \right)}}{7} + \frac{384 \sin^{5}{\left(x \right)}}{5} - \frac{256 \sin^{3}{\left(x \right)}}{3} + 64 \sin{\left(x \right)}

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

      ∫(−144cos⁡7(x)) dx=−144∫cos⁡7(x) dx\int \left(- 144 \cos^{7}{\left(x \right)}\right)\, dx = - 144 \int \cos^{7}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        cos⁡7(x)=(1−sin⁡2(x))3cos⁡(x)\cos^{7}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)}

      2. Rewrite the integrand:

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

      3. Integrate term-by-term:

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

          ∫(−sin⁡6(x)cos⁡(x)) dx=−∫sin⁡6(x)cos⁡(x) dx\int \left(- \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{6}{\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:

            ∫u6 du\int u^{6}\, du

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

              ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

            Now substitute uu back in:

            sin⁡7(x)7\frac{\sin^{7}{\left(x \right)}}{7}

          So, the result is: −sin⁡7(x)7- \frac{\sin^{7}{\left(x \right)}}{7}

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

          ∫3sin⁡4(x)cos⁡(x) dx=3∫sin⁡4(x)cos⁡(x) dx\int 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{4}{\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:

            ∫u4 du\int u^{4}\, du

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

              ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

            Now substitute uu back in:

            sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

          So, the result is: 3sin⁡5(x)5\frac{3 \sin^{5}{\left(x \right)}}{5}

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

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

        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⁡7(x)7+3sin⁡5(x)5−sin⁡3(x)+sin⁡(x)- \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} - \sin^{3}{\left(x \right)} + \sin{\left(x \right)}

      So, the result is: 144sin⁡7(x)7−432sin⁡5(x)5+144sin⁡3(x)−144sin⁡(x)\frac{144 \sin^{7}{\left(x \right)}}{7} - \frac{432 \sin^{5}{\left(x \right)}}{5} + 144 \sin^{3}{\left(x \right)} - 144 \sin{\left(x \right)}

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

      ∫116cos⁡5(x) dx=116∫cos⁡5(x) dx\int 116 \cos^{5}{\left(x \right)}\, dx = 116 \int \cos^{5}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        cos⁡5(x)=(1−sin⁡2(x))2cos⁡(x)\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}

      2. Rewrite the integrand:

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

      3. Integrate term-by-term:

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

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

          ∫u4 du\int u^{4}\, du

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

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          Now substitute uu back in:

          sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

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

          ∫(−2sin⁡2(x)cos⁡(x)) dx=−2∫sin⁡2(x)cos⁡(x) dx\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \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: −2sin⁡3(x)3- \frac{2 \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⁡5(x)5−2sin⁡3(x)3+sin⁡(x)\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

      So, the result is: 116sin⁡5(x)5−232sin⁡3(x)3+116sin⁡(x)\frac{116 \sin^{5}{\left(x \right)}}{5} - \frac{232 \sin^{3}{\left(x \right)}}{3} + 116 \sin{\left(x \right)}

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

      ∫(−40cos⁡3(x)) dx=−40∫cos⁡3(x) dx\int \left(- 40 \cos^{3}{\left(x \right)}\right)\, dx = - 40 \int \cos^{3}{\left(x \right)}\, dx

      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)}

      So, the result is: 40sin⁡3(x)3−40sin⁡(x)\frac{40 \sin^{3}{\left(x \right)}}{3} - 40 \sin{\left(x \right)}

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

      ∫5cos⁡(x) dx=5∫cos⁡(x) dx\int 5 \cos{\left(x \right)}\, dx = 5 \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: 5sin⁡(x)5 \sin{\left(x \right)}

    The result is: 64sin⁡9(x)9−16sin⁡7(x)+68sin⁡5(x)5−16sin⁡3(x)3+sin⁡(x)\frac{64 \sin^{9}{\left(x \right)}}{9} - 16 \sin^{7}{\left(x \right)} + \frac{68 \sin^{5}{\left(x \right)}}{5} - \frac{16 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}

  3. Now simplify:

    (320sin⁡8(x)−720sin⁡6(x)+612sin⁡4(x)−240sin⁡2(x)+45)sin⁡(x)45\frac{\left(320 \sin^{8}{\left(x \right)} - 720 \sin^{6}{\left(x \right)} + 612 \sin^{4}{\left(x \right)} - 240 \sin^{2}{\left(x \right)} + 45\right) \sin{\left(x \right)}}{45}

  4. Add the constant of integration:

    (320sin⁡8(x)−720sin⁡6(x)+612sin⁡4(x)−240sin⁡2(x)+45)sin⁡(x)45+constant\frac{\left(320 \sin^{8}{\left(x \right)} - 720 \sin^{6}{\left(x \right)} + 612 \sin^{4}{\left(x \right)} - 240 \sin^{2}{\left(x \right)} + 45\right) \sin{\left(x \right)}}{45}+ \mathrm{constant}

The answer is:

(320sin⁡8(x)−720sin⁡6(x)+612sin⁡4(x)−240sin⁡2(x)+45)sin⁡(x)45+constant\frac{\left(320 \sin^{8}{\left(x \right)} - 720 \sin^{6}{\left(x \right)} + 612 \sin^{4}{\left(x \right)} - 240 \sin^{2}{\left(x \right)} + 45\right) \sin{\left(x \right)}}{45}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                      
 |                                                3            9            5            
 |             2                     7      16*sin (x)   64*sin (x)   68*sin (x)         
 | cos(5*x)*cos (2*x) dx = C - 16*sin (x) - ---------- + ---------- + ---------- + sin(x)
 |                                              3            9            5              
/
∫cos⁡2(2x)cos⁡(5x) dx=C+64sin⁡9(x)9−16sin⁡7(x)+68sin⁡5(x)5−16sin⁡3(x)3+sin⁡(x)\int \cos^{2}{\left(2 x \right)} \cos{\left(5 x \right)}\, dx = C + \frac{64 \sin^{9}{\left(x \right)}}{9} - 16 \sin^{7}{\left(x \right)} + \frac{68 \sin^{5}{\left(x \right)}}{5} - \frac{16 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}
Simplify
The graph
1.00001.01001.00101.00201.00301.00401.00501.00601.00701.00801.00900.000.20
Plot the graph f(x)
The answer [src] 
0
00 
=
= 
0
00 
0
Numerical answer [src] 
0.0
0.0

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

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