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sin(x)^10
Solving integrals/ sin(x)^10

Integral of sin(x)^10 dx

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∫01sin⁡10(x) dx\int\limits_{0}^{1} \sin^{10}{\left(x \right)}\, dx 
Integral(sin(x)^10, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡10(x)=(12−cos⁡(2x)2)5\sin^{10}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{5}

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (12−cos⁡(2x)2)5=−cos⁡5(2x)32+5cos⁡4(2x)32−5cos⁡3(2x)16+5cos⁡2(2x)16−5cos⁡(2x)32+132\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{5} = - \frac{\cos^{5}{\left(2 x \right)}}{32} + \frac{5 \cos^{4}{\left(2 x \right)}}{32} - \frac{5 \cos^{3}{\left(2 x \right)}}{16} + \frac{5 \cos^{2}{\left(2 x \right)}}{16} - \frac{5 \cos{\left(2 x \right)}}{32} + \frac{1}{32}

    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(2x)32) dx=−∫cos⁡5(2x) dx32\int \left(- \frac{\cos^{5}{\left(2 x \right)}}{32}\right)\, dx = - \frac{\int \cos^{5}{\left(2 x \right)}\, dx}{32}

        1. Rewrite the integrand:

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

        2. There are multiple ways to do this integral.

          Method #1

          1. Let u=2xu = 2 x.

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

            ∫(sin⁡4(u)cos⁡(u)2−sin⁡2(u)cos⁡(u)+cos⁡(u)2) du\int \left(\frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2} - \sin^{2}{\left(u \right)} \cos{\left(u \right)} + \frac{\cos{\left(u \right)}}{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:

                ∫sin⁡4(u)cos⁡(u)2 du=∫sin⁡4(u)cos⁡(u) du2\int \frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2}\, du = \frac{\int \sin^{4}{\left(u \right)} \cos{\left(u \right)}\, du}{2}

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

                  Then let du=cos⁡(u)dudu = \cos{\left(u \right)} du 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(u)5\frac{\sin^{5}{\left(u \right)}}{5}

                So, the result is: sin⁡5(u)10\frac{\sin^{5}{\left(u \right)}}{10}

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

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

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

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

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

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

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

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

              The result is: sin⁡5(u)10−sin⁡3(u)3+sin⁡(u)2\frac{\sin^{5}{\left(u \right)}}{10} - \frac{\sin^{3}{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{2}

            Now substitute uu back in:

            sin⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}

          Method #2

          1. Rewrite the integrand:

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

          2. Integrate term-by-term:

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

              Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute du2\frac{du}{2}:

              ∫u44 du\int \frac{u^{4}}{4}\, du

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

                ∫u42 du=∫u4 du2\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}

                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: u510\frac{u^{5}}{10}

              Now substitute uu back in:

              sin⁡5(2x)10\frac{\sin^{5}{\left(2 x \right)}}{10}

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

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

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

                Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute du2\frac{du}{2}:

                ∫u24 du\int \frac{u^{2}}{4}\, du

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

                  ∫u22 du=∫u2 du2\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}

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

                    ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

                  So, the result is: u36\frac{u^{3}}{6}

                Now substitute uu back in:

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

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

            1. Let u=2xu = 2 x.

              Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

              ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

              Now substitute uu back in:

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

            The result is: sin⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}

          Method #3

          1. Rewrite the integrand:

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

          2. Integrate term-by-term:

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

              Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute du2\frac{du}{2}:

              ∫u44 du\int \frac{u^{4}}{4}\, du

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

                ∫u42 du=∫u4 du2\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}

                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: u510\frac{u^{5}}{10}

              Now substitute uu back in:

              sin⁡5(2x)10\frac{\sin^{5}{\left(2 x \right)}}{10}

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

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

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

                Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute du2\frac{du}{2}:

                ∫u24 du\int \frac{u^{2}}{4}\, du

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

                  ∫u22 du=∫u2 du2\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}

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

                    ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

                  So, the result is: u36\frac{u^{3}}{6}

                Now substitute uu back in:

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

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

            1. Let u=2xu = 2 x.

              Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

              ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

              Now substitute uu back in:

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

            The result is: sin⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡5(2x)320+sin⁡3(2x)96−sin⁡(2x)64- \frac{\sin^{5}{\left(2 x \right)}}{320} + \frac{\sin^{3}{\left(2 x \right)}}{96} - \frac{\sin{\left(2 x \right)}}{64}

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

        ∫5cos⁡4(2x)32 dx=5∫cos⁡4(2x) dx32\int \frac{5 \cos^{4}{\left(2 x \right)}}{32}\, dx = \frac{5 \int \cos^{4}{\left(2 x \right)}\, dx}{32}

        1. Rewrite the integrand:

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

        2. Rewrite the integrand:

          (cos⁡(4x)2+12)2=cos⁡2(4x)4+cos⁡(4x)2+14\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{4}

        3. Integrate term-by-term:

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

            ∫cos⁡2(4x)4 dx=∫cos⁡2(4x) dx4\int \frac{\cos^{2}{\left(4 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{4}

            1. Rewrite the integrand:

              cos⁡2(4x)=cos⁡(8x)2+12\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 x \right)}}{2} + \frac{1}{2}

            2. Integrate term-by-term:

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

                ∫cos⁡(8x)2 dx=∫cos⁡(8x) dx2\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}

                1. Let u=8xu = 8 x.

                  Then let du=8dxdu = 8 dx and substitute du8\frac{du}{8}:

                  ∫cos⁡(u)64 du\int \frac{\cos{\left(u \right)}}{64}\, du

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

                    ∫cos⁡(u)8 du=∫cos⁡(u) du8\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}

                    1. The integral of cosine is sine:

                      ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                    So, the result is: sin⁡(u)8\frac{\sin{\left(u \right)}}{8}

                  Now substitute uu back in:

                  sin⁡(8x)8\frac{\sin{\left(8 x \right)}}{8}

                So, the result is: sin⁡(8x)16\frac{\sin{\left(8 x \right)}}{16}

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

                ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

              The result is: x2+sin⁡(8x)16\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}

            So, the result is: x8+sin⁡(8x)64\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64}

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

            ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

            1. Let u=4xu = 4 x.

              Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

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

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

                ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

              Now substitute uu back in:

              sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

            So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

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

            ∫14 dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

          The result is: 3x8+sin⁡(4x)8+sin⁡(8x)64\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}

        So, the result is: 15x256+5sin⁡(4x)256+5sin⁡(8x)2048\frac{15 x}{256} + \frac{5 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}

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

        ∫(−5cos⁡3(2x)16) dx=−5∫cos⁡3(2x) dx16\int \left(- \frac{5 \cos^{3}{\left(2 x \right)}}{16}\right)\, dx = - \frac{5 \int \cos^{3}{\left(2 x \right)}\, dx}{16}

        1. Rewrite the integrand:

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

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

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

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

          1. Integrate term-by-term:

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

              ∫12 du=u2\int \frac{1}{2}\, du = \frac{u}{2}

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

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

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

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: −u36- \frac{u^{3}}{6}

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

          Now substitute uu back in:

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

        So, the result is: 5sin⁡3(2x)96−5sin⁡(2x)32\frac{5 \sin^{3}{\left(2 x \right)}}{96} - \frac{5 \sin{\left(2 x \right)}}{32}

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

        ∫5cos⁡2(2x)16 dx=5∫cos⁡2(2x) dx16\int \frac{5 \cos^{2}{\left(2 x \right)}}{16}\, dx = \frac{5 \int \cos^{2}{\left(2 x \right)}\, dx}{16}

        1. Rewrite the integrand:

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

        2. Integrate term-by-term:

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

            ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

            1. Let u=4xu = 4 x.

              Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

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

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

                ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

              Now substitute uu back in:

              sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

            So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

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

            ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

          The result is: x2+sin⁡(4x)8\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}

        So, the result is: 5x32+5sin⁡(4x)128\frac{5 x}{32} + \frac{5 \sin{\left(4 x \right)}}{128}

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

        ∫(−5cos⁡(2x)32) dx=−5∫cos⁡(2x) dx32\int \left(- \frac{5 \cos{\left(2 x \right)}}{32}\right)\, dx = - \frac{5 \int \cos{\left(2 x \right)}\, dx}{32}

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

            1. The integral of cosine is sine:

              ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

          Now substitute uu back in:

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

        So, the result is: −5sin⁡(2x)64- \frac{5 \sin{\left(2 x \right)}}{64}

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

        ∫132 dx=x32\int \frac{1}{32}\, dx = \frac{x}{32}

      The result is: 63x256−sin⁡5(2x)320+sin⁡3(2x)16−sin⁡(2x)4+15sin⁡(4x)256+5sin⁡(8x)2048\frac{63 x}{256} - \frac{\sin^{5}{\left(2 x \right)}}{320} + \frac{\sin^{3}{\left(2 x \right)}}{16} - \frac{\sin{\left(2 x \right)}}{4} + \frac{15 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}

    Method #2

    1. Rewrite the integrand:

      (12−cos⁡(2x)2)5=−cos⁡5(2x)32+5cos⁡4(2x)32−5cos⁡3(2x)16+5cos⁡2(2x)16−5cos⁡(2x)32+132\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{5} = - \frac{\cos^{5}{\left(2 x \right)}}{32} + \frac{5 \cos^{4}{\left(2 x \right)}}{32} - \frac{5 \cos^{3}{\left(2 x \right)}}{16} + \frac{5 \cos^{2}{\left(2 x \right)}}{16} - \frac{5 \cos{\left(2 x \right)}}{32} + \frac{1}{32}

    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(2x)32) dx=−∫cos⁡5(2x) dx32\int \left(- \frac{\cos^{5}{\left(2 x \right)}}{32}\right)\, dx = - \frac{\int \cos^{5}{\left(2 x \right)}\, dx}{32}

        1. Rewrite the integrand:

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

        2. Let u=2xu = 2 x.

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

          ∫(sin⁡4(u)cos⁡(u)2−sin⁡2(u)cos⁡(u)+cos⁡(u)2) du\int \left(\frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2} - \sin^{2}{\left(u \right)} \cos{\left(u \right)} + \frac{\cos{\left(u \right)}}{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:

              ∫sin⁡4(u)cos⁡(u)2 du=∫sin⁡4(u)cos⁡(u) du2\int \frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2}\, du = \frac{\int \sin^{4}{\left(u \right)} \cos{\left(u \right)}\, du}{2}

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

                Then let du=cos⁡(u)dudu = \cos{\left(u \right)} du 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(u)5\frac{\sin^{5}{\left(u \right)}}{5}

              So, the result is: sin⁡5(u)10\frac{\sin^{5}{\left(u \right)}}{10}

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

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

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

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

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

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

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

              1. The integral of cosine is sine:

                ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

            The result is: sin⁡5(u)10−sin⁡3(u)3+sin⁡(u)2\frac{\sin^{5}{\left(u \right)}}{10} - \frac{\sin^{3}{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡5(2x)320+sin⁡3(2x)96−sin⁡(2x)64- \frac{\sin^{5}{\left(2 x \right)}}{320} + \frac{\sin^{3}{\left(2 x \right)}}{96} - \frac{\sin{\left(2 x \right)}}{64}

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

        ∫5cos⁡4(2x)32 dx=5∫cos⁡4(2x) dx32\int \frac{5 \cos^{4}{\left(2 x \right)}}{32}\, dx = \frac{5 \int \cos^{4}{\left(2 x \right)}\, dx}{32}

        1. Rewrite the integrand:

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

        2. Rewrite the integrand:

          (cos⁡(4x)2+12)2=cos⁡2(4x)4+cos⁡(4x)2+14\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{4}

        3. Integrate term-by-term:

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

            ∫cos⁡2(4x)4 dx=∫cos⁡2(4x) dx4\int \frac{\cos^{2}{\left(4 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{4}

            1. Rewrite the integrand:

              cos⁡2(4x)=cos⁡(8x)2+12\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 x \right)}}{2} + \frac{1}{2}

            2. Integrate term-by-term:

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

                ∫cos⁡(8x)2 dx=∫cos⁡(8x) dx2\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}

                1. Let u=8xu = 8 x.

                  Then let du=8dxdu = 8 dx and substitute du8\frac{du}{8}:

                  ∫cos⁡(u)64 du\int \frac{\cos{\left(u \right)}}{64}\, du

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

                    ∫cos⁡(u)8 du=∫cos⁡(u) du8\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}

                    1. The integral of cosine is sine:

                      ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                    So, the result is: sin⁡(u)8\frac{\sin{\left(u \right)}}{8}

                  Now substitute uu back in:

                  sin⁡(8x)8\frac{\sin{\left(8 x \right)}}{8}

                So, the result is: sin⁡(8x)16\frac{\sin{\left(8 x \right)}}{16}

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

                ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

              The result is: x2+sin⁡(8x)16\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}

            So, the result is: x8+sin⁡(8x)64\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64}

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

            ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

            1. Let u=4xu = 4 x.

              Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

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

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

                ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

              Now substitute uu back in:

              sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

            So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

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

            ∫14 dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

          The result is: 3x8+sin⁡(4x)8+sin⁡(8x)64\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}

        So, the result is: 15x256+5sin⁡(4x)256+5sin⁡(8x)2048\frac{15 x}{256} + \frac{5 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}

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

        ∫(−5cos⁡3(2x)16) dx=−5∫cos⁡3(2x) dx16\int \left(- \frac{5 \cos^{3}{\left(2 x \right)}}{16}\right)\, dx = - \frac{5 \int \cos^{3}{\left(2 x \right)}\, dx}{16}

        1. Rewrite the integrand:

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

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

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

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

          1. Integrate term-by-term:

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

              ∫12 du=u2\int \frac{1}{2}\, du = \frac{u}{2}

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

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

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

                ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

              So, the result is: −u36- \frac{u^{3}}{6}

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

          Now substitute uu back in:

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

        So, the result is: 5sin⁡3(2x)96−5sin⁡(2x)32\frac{5 \sin^{3}{\left(2 x \right)}}{96} - \frac{5 \sin{\left(2 x \right)}}{32}

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

        ∫5cos⁡2(2x)16 dx=5∫cos⁡2(2x) dx16\int \frac{5 \cos^{2}{\left(2 x \right)}}{16}\, dx = \frac{5 \int \cos^{2}{\left(2 x \right)}\, dx}{16}

        1. Rewrite the integrand:

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

        2. Integrate term-by-term:

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

            ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

            1. Let u=4xu = 4 x.

              Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

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

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

                ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

              Now substitute uu back in:

              sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

            So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

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

            ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

          The result is: x2+sin⁡(4x)8\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}

        So, the result is: 5x32+5sin⁡(4x)128\frac{5 x}{32} + \frac{5 \sin{\left(4 x \right)}}{128}

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

        ∫(−5cos⁡(2x)32) dx=−5∫cos⁡(2x) dx32\int \left(- \frac{5 \cos{\left(2 x \right)}}{32}\right)\, dx = - \frac{5 \int \cos{\left(2 x \right)}\, dx}{32}

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

            1. The integral of cosine is sine:

              ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

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

          Now substitute uu back in:

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

        So, the result is: −5sin⁡(2x)64- \frac{5 \sin{\left(2 x \right)}}{64}

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

        ∫132 dx=x32\int \frac{1}{32}\, dx = \frac{x}{32}

      The result is: 63x256−sin⁡5(2x)320+sin⁡3(2x)16−sin⁡(2x)4+15sin⁡(4x)256+5sin⁡(8x)2048\frac{63 x}{256} - \frac{\sin^{5}{\left(2 x \right)}}{320} + \frac{\sin^{3}{\left(2 x \right)}}{16} - \frac{\sin{\left(2 x \right)}}{4} + \frac{15 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}

  3. Add the constant of integration:

    63x256−sin⁡5(2x)320+sin⁡3(2x)16−sin⁡(2x)4+15sin⁡(4x)256+5sin⁡(8x)2048+constant\frac{63 x}{256} - \frac{\sin^{5}{\left(2 x \right)}}{320} + \frac{\sin^{3}{\left(2 x \right)}}{16} - \frac{\sin{\left(2 x \right)}}{4} + \frac{15 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}+ \mathrm{constant}

The answer is:

63x256−sin⁡5(2x)320+sin⁡3(2x)16−sin⁡(2x)4+15sin⁡(4x)256+5sin⁡(8x)2048+constant\frac{63 x}{256} - \frac{\sin^{5}{\left(2 x \right)}}{320} + \frac{\sin^{3}{\left(2 x \right)}}{16} - \frac{\sin{\left(2 x \right)}}{4} + \frac{15 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                    
 |                                 5           3                                       
 |    10             sin(2*x)   sin (2*x)   sin (2*x)   5*sin(8*x)   15*sin(4*x)   63*x
 | sin  (x) dx = C - -------- - --------- + --------- + ---------- + ----------- + ----
 |                      4          320          16         2048          256       256 
/
5 (sin⁡(8 x)2+4 x8+sin⁡(4 x)2+x)64+5 (sin⁡(4 x)2+2 x)32−sin⁡5(2 x)5−2 sin⁡3(2 x)3+sin⁡(2 x)32−5 (sin⁡(2 x)−sin⁡3(2 x)3)16−5 sin⁡(2 x)32+x162{{{{5\,\left({{{{\sin \left(8\,x\right)}\over{2}}+4\,x}\over{8}}+{{ \sin \left(4\,x\right)}\over{2}}+x\right)}\over{64}}+{{5\,\left({{ \sin \left(4\,x\right)}\over{2}}+2\,x\right)}\over{32}}-{{{{\sin ^5 \left(2\,x\right)}\over{5}}-{{2\,\sin ^3\left(2\,x\right)}\over{3}}+ \sin \left(2\,x\right)}\over{32}}-{{5\,\left(\sin \left(2\,x\right)- {{\sin ^3\left(2\,x\right)}\over{3}}\right)}\over{16}}-{{5\,\sin \left(2\,x\right)}\over{32}}+{{x}\over{16}}}\over{2}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.2
Plot the graph f(x)
The answer [src] 
                               3                   5                  7                9          
 63   63*cos(1)*sin(1)   21*sin (1)*cos(1)   21*sin (1)*cos(1)   9*sin (1)*cos(1)   sin (1)*cos(1)
--- - ---------------- - ----------------- - ----------------- - ---------------- - --------------
256         256                 128                 160                 80                10
25 sin⁡8+600 sin⁡4−32 sin⁡52+640 sin⁡32−2560 sin⁡2+252010240{{25\,\sin 8+600\,\sin 4-32\,\sin ^52+640\,\sin ^32-2560\,\sin 2+ 2520}\over{10240}} 
=
= 
                               3                   5                  7                9          
 63   63*cos(1)*sin(1)   21*sin (1)*cos(1)   21*sin (1)*cos(1)   9*sin (1)*cos(1)   sin (1)*cos(1)
--- - ---------------- - ----------------- - ----------------- - ---------------- - --------------
256         256                 128                 160                 80                10
−63sin⁡(1)cos⁡(1)256−21sin⁡3(1)cos⁡(1)128−21sin⁡5(1)cos⁡(1)160−9sin⁡7(1)cos⁡(1)80−sin⁡9(1)cos⁡(1)10+63256- \frac{63 \sin{\left(1 \right)} \cos{\left(1 \right)}}{256} - \frac{21 \sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{128} - \frac{21 \sin^{5}{\left(1 \right)} \cos{\left(1 \right)}}{160} - \frac{9 \sin^{7}{\left(1 \right)} \cos{\left(1 \right)}}{80} - \frac{\sin^{9}{\left(1 \right)} \cos{\left(1 \right)}}{10} + \frac{63}{256}
Simplify
Numerical answer [src] 
0.0218875224217298
0.0218875224217298
The graph
Integral of sin(x)^10 dx

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

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