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sin(5x)cos(2x)dx
Solving integrals/ sin(5x)cos(2x)dx

Integral of sin(5x)cos(2x)dx dx

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∫01sin⁡(5x)cos⁡(2x)1 dx\int\limits_{0}^{1} \sin{\left(5 x \right)} \cos{\left(2 x \right)} 1\, dx 
Integral(sin(5*x)*cos(2*x)*1, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡(5x)cos⁡(2x)1=32sin⁡5(x)cos⁡2(x)−16sin⁡5(x)−40sin⁡3(x)cos⁡2(x)+20sin⁡3(x)+10sin⁡(x)cos⁡2(x)−5sin⁡(x)\sin{\left(5 x \right)} \cos{\left(2 x \right)} 1 = 32 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)} - 16 \sin^{5}{\left(x \right)} - 40 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} + 20 \sin^{3}{\left(x \right)} + 10 \sin{\left(x \right)} \cos^{2}{\left(x \right)} - 5 \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:

      ∫32sin⁡5(x)cos⁡2(x) dx=32∫sin⁡5(x)cos⁡2(x) dx\int 32 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 32 \int \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

          ∫(−u6+2u4−u2) du\int \left(- u^{6} + 2 u^{4} - u^{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:

              ∫(−u6) du=−∫u6 du\int \left(- u^{6}\right)\, 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}

              So, the result is: −u77- \frac{u^{7}}{7}

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

              ∫2u4 du=2∫u4 du\int 2 u^{4}\, du = 2 \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}

              So, the result is: 2u55\frac{2 u^{5}}{5}

            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: −u77+2u55−u33- \frac{u^{7}}{7} + \frac{2 u^{5}}{5} - \frac{u^{3}}{3}

          Now substitute uu back in:

          −cos⁡7(x)7+2cos⁡5(x)5−cos⁡3(x)3- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

        Method #2

        1. Rewrite the integrand:

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

        2. Integrate term-by-term:

          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:

            ∫u6 du\int u^{6}\, du

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

              ∫(−u6) du=−∫u6 du\int \left(- u^{6}\right)\, 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}

              So, the result is: −u77- \frac{u^{7}}{7}

            Now substitute uu back in:

            −cos⁡7(x)7- \frac{\cos^{7}{\left(x \right)}}{7}

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

            ∫(−2sin⁡(x)cos⁡4(x)) dx=−2∫sin⁡(x)cos⁡4(x) dx\int \left(- 2 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)} \cos^{4}{\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:

              ∫u4 du\int u^{4}\, du

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

                ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, 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}

                So, the result is: −u55- \frac{u^{5}}{5}

              Now substitute uu back in:

              −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

            So, the result is: 2cos⁡5(x)5\frac{2 \cos^{5}{\left(x \right)}}{5}

          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}

          The result is: −cos⁡7(x)7+2cos⁡5(x)5−cos⁡3(x)3- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

        Method #3

        1. Rewrite the integrand:

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

        2. Integrate term-by-term:

          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:

            ∫u6 du\int u^{6}\, du

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

              ∫(−u6) du=−∫u6 du\int \left(- u^{6}\right)\, 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}

              So, the result is: −u77- \frac{u^{7}}{7}

            Now substitute uu back in:

            −cos⁡7(x)7- \frac{\cos^{7}{\left(x \right)}}{7}

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

            ∫(−2sin⁡(x)cos⁡4(x)) dx=−2∫sin⁡(x)cos⁡4(x) dx\int \left(- 2 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)} \cos^{4}{\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:

              ∫u4 du\int u^{4}\, du

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

                ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, 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}

                So, the result is: −u55- \frac{u^{5}}{5}

              Now substitute uu back in:

              −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

            So, the result is: 2cos⁡5(x)5\frac{2 \cos^{5}{\left(x \right)}}{5}

          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}

          The result is: −cos⁡7(x)7+2cos⁡5(x)5−cos⁡3(x)3- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: −32cos⁡7(x)7+64cos⁡5(x)5−32cos⁡3(x)3- \frac{32 \cos^{7}{\left(x \right)}}{7} + \frac{64 \cos^{5}{\left(x \right)}}{5} - \frac{32 \cos^{3}{\left(x \right)}}{3}

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

      ∫(−16sin⁡5(x)) dx=−16∫sin⁡5(x) dx\int \left(- 16 \sin^{5}{\left(x \right)}\right)\, dx = - 16 \int \sin^{5}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

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

      3. Integrate term-by-term:

        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:

          ∫u4 du\int u^{4}\, du

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

            ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, 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}

            So, the result is: −u55- \frac{u^{5}}{5}

          Now substitute uu back in:

          −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

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

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

      So, the result is: 16cos⁡5(x)5−32cos⁡3(x)3+16cos⁡(x)\frac{16 \cos^{5}{\left(x \right)}}{5} - \frac{32 \cos^{3}{\left(x \right)}}{3} + 16 \cos{\left(x \right)}

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

      ∫(−40sin⁡3(x)cos⁡2(x)) dx=−40∫sin⁡3(x)cos⁡2(x) dx\int \left(- 40 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 40 \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

        ∫(u4−u2) du\int \left(u^{4} - u^{2}\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:

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

          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: u55−u33\frac{u^{5}}{5} - \frac{u^{3}}{3}

        Now substitute uu back in:

        cos⁡5(x)5−cos⁡3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

      So, the result is: −8cos⁡5(x)+40cos⁡3(x)3- 8 \cos^{5}{\left(x \right)} + \frac{40 \cos^{3}{\left(x \right)}}{3}

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

      ∫20sin⁡3(x) dx=20∫sin⁡3(x) dx\int 20 \sin^{3}{\left(x \right)}\, dx = 20 \int \sin^{3}{\left(x \right)}\, dx

      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: 20cos⁡3(x)3−20cos⁡(x)\frac{20 \cos^{3}{\left(x \right)}}{3} - 20 \cos{\left(x \right)}

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

      ∫10sin⁡(x)cos⁡2(x) dx=10∫sin⁡(x)cos⁡2(x) dx\int 10 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 10 \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: −10cos⁡3(x)3- \frac{10 \cos^{3}{\left(x \right)}}{3}

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

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

      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: 5cos⁡(x)5 \cos{\left(x \right)}

    The result is: −32cos⁡7(x)7+8cos⁡5(x)−14cos⁡3(x)3+cos⁡(x)- \frac{32 \cos^{7}{\left(x \right)}}{7} + 8 \cos^{5}{\left(x \right)} - \frac{14 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}

  3. Now simplify:

    (−96cos⁡6(x)+168cos⁡4(x)−98cos⁡2(x)+21)cos⁡(x)21\frac{\left(- 96 \cos^{6}{\left(x \right)} + 168 \cos^{4}{\left(x \right)} - 98 \cos^{2}{\left(x \right)} + 21\right) \cos{\left(x \right)}}{21}

  4. Add the constant of integration:

    (−96cos⁡6(x)+168cos⁡4(x)−98cos⁡2(x)+21)cos⁡(x)21+constant\frac{\left(- 96 \cos^{6}{\left(x \right)} + 168 \cos^{4}{\left(x \right)} - 98 \cos^{2}{\left(x \right)} + 21\right) \cos{\left(x \right)}}{21}+ \mathrm{constant}

The answer is:

(−96cos⁡6(x)+168cos⁡4(x)−98cos⁡2(x)+21)cos⁡(x)21+constant\frac{\left(- 96 \cos^{6}{\left(x \right)} + 168 \cos^{4}{\left(x \right)} - 98 \cos^{2}{\left(x \right)} + 21\right) \cos{\left(x \right)}}{21}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                               7            3            
 |                                   5      32*cos (x)   14*cos (x)         
 | sin(5*x)*cos(2*x)*1 dx = C + 8*cos (x) - ---------- - ---------- + cos(x)
 |                                              7            3              
/
−cos⁡(7 x)14−cos⁡(3 x)6-{{\cos \left(7\,x\right)}\over{14}}-{{\cos \left(3\,x\right) }\over{6}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.901-1
Plot the graph f(x)
The answer [src] 
5    5*cos(2)*cos(5)   2*sin(2)*sin(5)
-- - --------------- - ---------------
21          21                21
521−3 cos⁡7+7 cos⁡342{{5}\over{21}}-{{3\,\cos 7+7\,\cos 3}\over{42}} 
=
= 
5    5*cos(2)*cos(5)   2*sin(2)*sin(5)
-- - --------------- - ---------------
21          21                21
−5cos⁡(2)cos⁡(5)21−2sin⁡(2)sin⁡(5)21+521- \frac{5 \cos{\left(2 \right)} \cos{\left(5 \right)}}{21} - \frac{2 \sin{\left(2 \right)} \sin{\left(5 \right)}}{21} + \frac{5}{21}
Simplify
Numerical answer [src] 
0.349243826504124
0.349243826504124
The graph
Integral of sin(5x)cos(2x)dx dx

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

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