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Solving integrals/ (sin^3)*((6x)*cos6x)

Integral of (sin^3)*((6x)*cos6x) dx

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∫00sin⁡3(x)6xcos⁡(6x) dx\int\limits_{0}^{0} \sin^{3}{\left(x \right)} 6 x \cos{\left(6 x \right)}\, dx 
Integral(sin(x)^3*6*x*cos(6*x), (x, 0, 0))
Detail solution

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

    ∫sin⁡3(x)6xcos⁡(6x) dx=6∫xsin⁡3(x)cos⁡(6x) dx\int \sin^{3}{\left(x \right)} 6 x \cos{\left(6 x \right)}\, dx = 6 \int x \sin^{3}{\left(x \right)} \cos{\left(6 x \right)}\, dx

    1. There are multiple ways to do this integral.

      Method #1

      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⁡(6x)\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos{\left(6 x \right)}.

        Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

        To find v(x)v{\left(x \right)}:

        1. Rewrite the integrand:

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

        2. Rewrite the integrand:

          (1−cos⁡2(x))sin⁡(x)cos⁡(6x)=−sin⁡(x)cos⁡2(x)cos⁡(6x)+sin⁡(x)cos⁡(6x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos{\left(6 x \right)} = - \sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)} + \sin{\left(x \right)} \cos{\left(6 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⁡(x)cos⁡2(x)cos⁡(6x)) dx=−∫sin⁡(x)cos⁡2(x)cos⁡(6x) dx\int \left(- \sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)}\, dx

            1. Rewrite the integrand:

              sin⁡(x)cos⁡2(x)cos⁡(6x)=32sin⁡(x)cos⁡8(x)−48sin⁡(x)cos⁡6(x)+18sin⁡(x)cos⁡4(x)−sin⁡(x)cos⁡2(x)\sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)} = 32 \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 48 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + 18 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - \sin{\left(x \right)} \cos^{2}{\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⁡(x)cos⁡8(x) dx=32∫sin⁡(x)cos⁡8(x) dx\int 32 \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx = 32 \int \sin{\left(x \right)} \cos^{8}{\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:

                  ∫u8 du\int u^{8}\, du

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

                    ∫(−u8) du=−∫u8 du\int \left(- u^{8}\right)\, 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}

                    So, the result is: −u99- \frac{u^{9}}{9}

                  Now substitute uu back in:

                  −cos⁡9(x)9- \frac{\cos^{9}{\left(x \right)}}{9}

                So, the result is: −32cos⁡9(x)9- \frac{32 \cos^{9}{\left(x \right)}}{9}

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

                ∫(−48sin⁡(x)cos⁡6(x)) dx=−48∫sin⁡(x)cos⁡6(x) dx\int \left(- 48 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 48 \int \sin{\left(x \right)} \cos^{6}{\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:

                  ∫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}

                So, the result is: 48cos⁡7(x)7\frac{48 \cos^{7}{\left(x \right)}}{7}

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

                ∫18sin⁡(x)cos⁡4(x) dx=18∫sin⁡(x)cos⁡4(x) dx\int 18 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 18 \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: −18cos⁡5(x)5- \frac{18 \cos^{5}{\left(x \right)}}{5}

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

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

              The result is: −32cos⁡9(x)9+48cos⁡7(x)7−18cos⁡5(x)5+cos⁡3(x)3- \frac{32 \cos^{9}{\left(x \right)}}{9} + \frac{48 \cos^{7}{\left(x \right)}}{7} - \frac{18 \cos^{5}{\left(x \right)}}{5} + \frac{\cos^{3}{\left(x \right)}}{3}

            So, the result is: 32cos⁡9(x)9−48cos⁡7(x)7+18cos⁡5(x)5−cos⁡3(x)3\frac{32 \cos^{9}{\left(x \right)}}{9} - \frac{48 \cos^{7}{\left(x \right)}}{7} + \frac{18 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

          1. Rewrite the integrand:

            sin⁡(x)cos⁡(6x)=32sin⁡(x)cos⁡6(x)−48sin⁡(x)cos⁡4(x)+18sin⁡(x)cos⁡2(x)−sin⁡(x)\sin{\left(x \right)} \cos{\left(6 x \right)} = 32 \sin{\left(x \right)} \cos^{6}{\left(x \right)} - 48 \sin{\left(x \right)} \cos^{4}{\left(x \right)} + 18 \sin{\left(x \right)} \cos^{2}{\left(x \right)} - \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⁡(x)cos⁡6(x) dx=32∫sin⁡(x)cos⁡6(x) dx\int 32 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx = 32 \int \sin{\left(x \right)} \cos^{6}{\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:

                ∫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}

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

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

              ∫(−48sin⁡(x)cos⁡4(x)) dx=−48∫sin⁡(x)cos⁡4(x) dx\int \left(- 48 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 48 \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: 48cos⁡5(x)5\frac{48 \cos^{5}{\left(x \right)}}{5}

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

              ∫18sin⁡(x)cos⁡2(x) dx=18∫sin⁡(x)cos⁡2(x) dx\int 18 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 18 \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: −6cos⁡3(x)- 6 \cos^{3}{\left(x \right)}

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

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

            The result is: −32cos⁡7(x)7+48cos⁡5(x)5−6cos⁡3(x)+cos⁡(x)- \frac{32 \cos^{7}{\left(x \right)}}{7} + \frac{48 \cos^{5}{\left(x \right)}}{5} - 6 \cos^{3}{\left(x \right)} + \cos{\left(x \right)}

          The result is: 32cos⁡9(x)9−80cos⁡7(x)7+66cos⁡5(x)5−19cos⁡3(x)3+cos⁡(x)\frac{32 \cos^{9}{\left(x \right)}}{9} - \frac{80 \cos^{7}{\left(x \right)}}{7} + \frac{66 \cos^{5}{\left(x \right)}}{5} - \frac{19 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}

        Now evaluate the sub-integral.

      2. Integrate term-by-term:

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

          ∫32cos⁡9(x)9 dx=32∫cos⁡9(x) dx9\int \frac{32 \cos^{9}{\left(x \right)}}{9}\, dx = \frac{32 \int \cos^{9}{\left(x \right)}\, dx}{9}

          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. 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)}

          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:

              ∫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: 32sin⁡9(x)81−128sin⁡7(x)63+64sin⁡5(x)15−128sin⁡3(x)27+32sin⁡(x)9\frac{32 \sin^{9}{\left(x \right)}}{81} - \frac{128 \sin^{7}{\left(x \right)}}{63} + \frac{64 \sin^{5}{\left(x \right)}}{15} - \frac{128 \sin^{3}{\left(x \right)}}{27} + \frac{32 \sin{\left(x \right)}}{9}

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

          ∫(−80cos⁡7(x)7) dx=−80∫cos⁡7(x) dx7\int \left(- \frac{80 \cos^{7}{\left(x \right)}}{7}\right)\, dx = - \frac{80 \int \cos^{7}{\left(x \right)}\, dx}{7}

          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: 80sin⁡7(x)49−48sin⁡5(x)7+80sin⁡3(x)7−80sin⁡(x)7\frac{80 \sin^{7}{\left(x \right)}}{49} - \frac{48 \sin^{5}{\left(x \right)}}{7} + \frac{80 \sin^{3}{\left(x \right)}}{7} - \frac{80 \sin{\left(x \right)}}{7}

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

          ∫66cos⁡5(x)5 dx=66∫cos⁡5(x) dx5\int \frac{66 \cos^{5}{\left(x \right)}}{5}\, dx = \frac{66 \int \cos^{5}{\left(x \right)}\, dx}{5}

          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: 66sin⁡5(x)25−44sin⁡3(x)5+66sin⁡(x)5\frac{66 \sin^{5}{\left(x \right)}}{25} - \frac{44 \sin^{3}{\left(x \right)}}{5} + \frac{66 \sin{\left(x \right)}}{5}

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

          ∫(−19cos⁡3(x)3) dx=−19∫cos⁡3(x) dx3\int \left(- \frac{19 \cos^{3}{\left(x \right)}}{3}\right)\, dx = - \frac{19 \int \cos^{3}{\left(x \right)}\, dx}{3}

          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: 19sin⁡3(x)9−19sin⁡(x)3\frac{19 \sin^{3}{\left(x \right)}}{9} - \frac{19 \sin{\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: 32sin⁡9(x)81−176sin⁡7(x)441+26sin⁡5(x)525−sin⁡3(x)945−2sin⁡(x)315\frac{32 \sin^{9}{\left(x \right)}}{81} - \frac{176 \sin^{7}{\left(x \right)}}{441} + \frac{26 \sin^{5}{\left(x \right)}}{525} - \frac{\sin^{3}{\left(x \right)}}{945} - \frac{2 \sin{\left(x \right)}}{315}

      Method #2

      1. Rewrite the integrand:

        xsin⁡3(x)cos⁡(6x)=32xsin⁡3(x)cos⁡6(x)−48xsin⁡3(x)cos⁡4(x)+18xsin⁡3(x)cos⁡2(x)−xsin⁡3(x)x \sin^{3}{\left(x \right)} \cos{\left(6 x \right)} = 32 x \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)} - 48 x \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} + 18 x \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} - x \sin^{3}{\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:

          ∫32xsin⁡3(x)cos⁡6(x) dx=32∫xsin⁡3(x)cos⁡6(x) dx\int 32 x \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}\, dx = 32 \int x \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}\, dx

          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⁡6(x)\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}.

            Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

            To find v(x)v{\left(x \right)}:

            1. Rewrite the integrand:

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

              ∫(u8−u6) du\int \left(u^{8} - u^{6}\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:

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

                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}

                The result is: u99−u77\frac{u^{9}}{9} - \frac{u^{7}}{7}

              Now substitute uu back in:

              cos⁡9(x)9−cos⁡7(x)7\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}

            Now evaluate the sub-integral.

          2. Integrate term-by-term:

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

              ∫cos⁡9(x)9 dx=∫cos⁡9(x) dx9\int \frac{\cos^{9}{\left(x \right)}}{9}\, dx = \frac{\int \cos^{9}{\left(x \right)}\, dx}{9}

              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. 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)}

              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:

                  ∫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: sin⁡9(x)81−4sin⁡7(x)63+2sin⁡5(x)15−4sin⁡3(x)27+sin⁡(x)9\frac{\sin^{9}{\left(x \right)}}{81} - \frac{4 \sin^{7}{\left(x \right)}}{63} + \frac{2 \sin^{5}{\left(x \right)}}{15} - \frac{4 \sin^{3}{\left(x \right)}}{27} + \frac{\sin{\left(x \right)}}{9}

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

              ∫(−cos⁡7(x)7) dx=−∫cos⁡7(x) dx7\int \left(- \frac{\cos^{7}{\left(x \right)}}{7}\right)\, dx = - \frac{\int \cos^{7}{\left(x \right)}\, dx}{7}

              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: sin⁡7(x)49−3sin⁡5(x)35+sin⁡3(x)7−sin⁡(x)7\frac{\sin^{7}{\left(x \right)}}{49} - \frac{3 \sin^{5}{\left(x \right)}}{35} + \frac{\sin^{3}{\left(x \right)}}{7} - \frac{\sin{\left(x \right)}}{7}

            The result is: sin⁡9(x)81−19sin⁡7(x)441+sin⁡5(x)21−sin⁡3(x)189−2sin⁡(x)63\frac{\sin^{9}{\left(x \right)}}{81} - \frac{19 \sin^{7}{\left(x \right)}}{441} + \frac{\sin^{5}{\left(x \right)}}{21} - \frac{\sin^{3}{\left(x \right)}}{189} - \frac{2 \sin{\left(x \right)}}{63}

          So, the result is: 32x(cos⁡9(x)9−cos⁡7(x)7)−32sin⁡9(x)81+608sin⁡7(x)441−32sin⁡5(x)21+32sin⁡3(x)189+64sin⁡(x)6332 x \left(\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}\right) - \frac{32 \sin^{9}{\left(x \right)}}{81} + \frac{608 \sin^{7}{\left(x \right)}}{441} - \frac{32 \sin^{5}{\left(x \right)}}{21} + \frac{32 \sin^{3}{\left(x \right)}}{189} + \frac{64 \sin{\left(x \right)}}{63}

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

          ∫(−48xsin⁡3(x)cos⁡4(x)) dx=−48∫xsin⁡3(x)cos⁡4(x) dx\int \left(- 48 x \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 48 \int x \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

          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⁡4(x)\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}.

            Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

            To find v(x)v{\left(x \right)}:

            1. Rewrite the integrand:

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

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

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

                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}

                The result is: u77−u55\frac{u^{7}}{7} - \frac{u^{5}}{5}

              Now substitute uu back in:

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

            Now evaluate the sub-integral.

          2. Integrate term-by-term:

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

              ∫cos⁡7(x)7 dx=∫cos⁡7(x) dx7\int \frac{\cos^{7}{\left(x \right)}}{7}\, dx = \frac{\int \cos^{7}{\left(x \right)}\, dx}{7}

              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: −sin⁡7(x)49+3sin⁡5(x)35−sin⁡3(x)7+sin⁡(x)7- \frac{\sin^{7}{\left(x \right)}}{49} + \frac{3 \sin^{5}{\left(x \right)}}{35} - \frac{\sin^{3}{\left(x \right)}}{7} + \frac{\sin{\left(x \right)}}{7}

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

              ∫(−cos⁡5(x)5) dx=−∫cos⁡5(x) dx5\int \left(- \frac{\cos^{5}{\left(x \right)}}{5}\right)\, dx = - \frac{\int \cos^{5}{\left(x \right)}\, dx}{5}

              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: −sin⁡5(x)25+2sin⁡3(x)15−sin⁡(x)5- \frac{\sin^{5}{\left(x \right)}}{25} + \frac{2 \sin^{3}{\left(x \right)}}{15} - \frac{\sin{\left(x \right)}}{5}

            The result is: −sin⁡7(x)49+8sin⁡5(x)175−sin⁡3(x)105−2sin⁡(x)35- \frac{\sin^{7}{\left(x \right)}}{49} + \frac{8 \sin^{5}{\left(x \right)}}{175} - \frac{\sin^{3}{\left(x \right)}}{105} - \frac{2 \sin{\left(x \right)}}{35}

          So, the result is: −48x(cos⁡7(x)7−cos⁡5(x)5)−48sin⁡7(x)49+384sin⁡5(x)175−16sin⁡3(x)35−96sin⁡(x)35- 48 x \left(\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}\right) - \frac{48 \sin^{7}{\left(x \right)}}{49} + \frac{384 \sin^{5}{\left(x \right)}}{175} - \frac{16 \sin^{3}{\left(x \right)}}{35} - \frac{96 \sin{\left(x \right)}}{35}

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

          ∫18xsin⁡3(x)cos⁡2(x) dx=18∫xsin⁡3(x)cos⁡2(x) dx\int 18 x \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 18 \int x \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

          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⁡2(x)\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}.

            Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

            To find v(x)v{\left(x \right)}:

            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}

            Now evaluate the sub-integral.

          2. Integrate term-by-term:

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

              ∫cos⁡5(x)5 dx=∫cos⁡5(x) dx5\int \frac{\cos^{5}{\left(x \right)}}{5}\, dx = \frac{\int \cos^{5}{\left(x \right)}\, dx}{5}

              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: sin⁡5(x)25−2sin⁡3(x)15+sin⁡(x)5\frac{\sin^{5}{\left(x \right)}}{25} - \frac{2 \sin^{3}{\left(x \right)}}{15} + \frac{\sin{\left(x \right)}}{5}

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

              ∫(−cos⁡3(x)3) dx=−∫cos⁡3(x) dx3\int \left(- \frac{\cos^{3}{\left(x \right)}}{3}\right)\, dx = - \frac{\int \cos^{3}{\left(x \right)}\, dx}{3}

              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: sin⁡3(x)9−sin⁡(x)3\frac{\sin^{3}{\left(x \right)}}{9} - \frac{\sin{\left(x \right)}}{3}

            The result is: sin⁡5(x)25−sin⁡3(x)45−2sin⁡(x)15\frac{\sin^{5}{\left(x \right)}}{25} - \frac{\sin^{3}{\left(x \right)}}{45} - \frac{2 \sin{\left(x \right)}}{15}

          So, the result is: 18x(cos⁡5(x)5−cos⁡3(x)3)−18sin⁡5(x)25+2sin⁡3(x)5+12sin⁡(x)518 x \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) - \frac{18 \sin^{5}{\left(x \right)}}{25} + \frac{2 \sin^{3}{\left(x \right)}}{5} + \frac{12 \sin{\left(x \right)}}{5}

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

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

          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)\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)}.

            Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

            To find v(x)v{\left(x \right)}:

            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)}

            Now evaluate the sub-integral.

          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(x)3 dx=∫cos⁡3(x) dx3\int \frac{\cos^{3}{\left(x \right)}}{3}\, dx = \frac{\int \cos^{3}{\left(x \right)}\, dx}{3}

              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: −sin⁡3(x)9+sin⁡(x)3- \frac{\sin^{3}{\left(x \right)}}{9} + \frac{\sin{\left(x \right)}}{3}

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

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

            The result is: −sin⁡3(x)9−2sin⁡(x)3- \frac{\sin^{3}{\left(x \right)}}{9} - \frac{2 \sin{\left(x \right)}}{3}

          So, the result is: −x(cos⁡3(x)3−cos⁡(x))−sin⁡3(x)9−2sin⁡(x)3- x \left(\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) - \frac{\sin^{3}{\left(x \right)}}{9} - \frac{2 \sin{\left(x \right)}}{3}

        The result is: −x(cos⁡3(x)3−cos⁡(x))+18x(cos⁡5(x)5−cos⁡3(x)3)−48x(cos⁡7(x)7−cos⁡5(x)5)+32x(cos⁡9(x)9−cos⁡7(x)7)−32sin⁡9(x)81+176sin⁡7(x)441−26sin⁡5(x)525+sin⁡3(x)945+2sin⁡(x)315- x \left(\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) + 18 x \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) - 48 x \left(\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}\right) + 32 x \left(\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}\right) - \frac{32 \sin^{9}{\left(x \right)}}{81} + \frac{176 \sin^{7}{\left(x \right)}}{441} - \frac{26 \sin^{5}{\left(x \right)}}{525} + \frac{\sin^{3}{\left(x \right)}}{945} + \frac{2 \sin{\left(x \right)}}{315}

    So, the result is: 6x(32cos⁡9(x)9−80cos⁡7(x)7+66cos⁡5(x)5−19cos⁡3(x)3+cos⁡(x))−64sin⁡9(x)27+352sin⁡7(x)147−52sin⁡5(x)175+2sin⁡3(x)315+4sin⁡(x)1056 x \left(\frac{32 \cos^{9}{\left(x \right)}}{9} - \frac{80 \cos^{7}{\left(x \right)}}{7} + \frac{66 \cos^{5}{\left(x \right)}}{5} - \frac{19 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}\right) - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}

  2. Now simplify:

    2x(1120cos⁡8(x)−3600cos⁡6(x)+4158cos⁡4(x)−1995cos⁡2(x)+315)cos⁡(x)105−64sin⁡9(x)27+352sin⁡7(x)147−52sin⁡5(x)175+2sin⁡3(x)315+4sin⁡(x)105\frac{2 x \left(1120 \cos^{8}{\left(x \right)} - 3600 \cos^{6}{\left(x \right)} + 4158 \cos^{4}{\left(x \right)} - 1995 \cos^{2}{\left(x \right)} + 315\right) \cos{\left(x \right)}}{105} - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}

  3. Add the constant of integration:

    2x(1120cos⁡8(x)−3600cos⁡6(x)+4158cos⁡4(x)−1995cos⁡2(x)+315)cos⁡(x)105−64sin⁡9(x)27+352sin⁡7(x)147−52sin⁡5(x)175+2sin⁡3(x)315+4sin⁡(x)105+constant\frac{2 x \left(1120 \cos^{8}{\left(x \right)} - 3600 \cos^{6}{\left(x \right)} + 4158 \cos^{4}{\left(x \right)} - 1995 \cos^{2}{\left(x \right)} + 315\right) \cos{\left(x \right)}}{105} - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}+ \mathrm{constant}

The answer is:

2x(1120cos⁡8(x)−3600cos⁡6(x)+4158cos⁡4(x)−1995cos⁡2(x)+315)cos⁡(x)105−64sin⁡9(x)27+352sin⁡7(x)147−52sin⁡5(x)175+2sin⁡3(x)315+4sin⁡(x)105+constant\frac{2 x \left(1120 \cos^{8}{\left(x \right)} - 3600 \cos^{6}{\left(x \right)} + 4158 \cos^{4}{\left(x \right)} - 1995 \cos^{2}{\left(x \right)} + 315\right) \cos{\left(x \right)}}{105} - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                                                                                               
 |                                     9            5           3                        7          /        7            3            9            5            \
 |    3                          64*sin (x)   52*sin (x)   2*sin (x)   4*sin(x)   352*sin (x)       |  80*cos (x)   19*cos (x)   32*cos (x)   66*cos (x)         |
 | sin (x)*6*x*cos(6*x) dx = C - ---------- - ---------- + --------- + -------- + ----------- + 6*x*|- ---------- - ---------- + ---------- + ---------- + cos(x)|
 |                                   27          175          315        105          147           \      7            3            9            5              /
/
−1225 sin⁡(9 x)−11025 x cos⁡(9 x)−6075 sin⁡(7 x)+42525 x cos⁡(7 x)+11907 sin⁡(5 x)−59535 x cos⁡(5 x)−11025 sin⁡(3 x)+33075 x cos⁡(3 x)132300-{{1225\,\sin \left(9\,x\right)-11025\,x\,\cos \left(9\,x\right)- 6075\,\sin \left(7\,x\right)+42525\,x\,\cos \left(7\,x\right)+11907 \,\sin \left(5\,x\right)-59535\,x\,\cos \left(5\,x\right)-11025\, \sin \left(3\,x\right)+33075\,x\,\cos \left(3\,x\right)}\over{132300 }}
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    Use the examples entering the upper and lower limits of integration.

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