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cos(7x)cos(3x)
Solving integrals/ cos(7x)cos(3x)

Integral of cos(7x)cos(3x) dx

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∫01cos⁡(3x)cos⁡(7x) dx\int\limits_{0}^{1} \cos{\left(3 x \right)} \cos{\left(7 x \right)}\, dx 
Integral(cos(7*x)*cos(3*x), (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    cos⁡(3x)cos⁡(7x)=256cos⁡10(x)−640cos⁡8(x)+560cos⁡6(x)−196cos⁡4(x)+21cos⁡2(x)\cos{\left(3 x \right)} \cos{\left(7 x \right)} = 256 \cos^{10}{\left(x \right)} - 640 \cos^{8}{\left(x \right)} + 560 \cos^{6}{\left(x \right)} - 196 \cos^{4}{\left(x \right)} + 21 \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:

      ∫256cos⁡10(x) dx=256∫cos⁡10(x) dx\int 256 \cos^{10}{\left(x \right)}\, dx = 256 \int \cos^{10}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. There are multiple ways to do this integral.

        Method #1

        1. Rewrite the integrand:

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

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

      So, the result is: 63x+4sin⁡5(2x)5−16sin⁡3(2x)+64sin⁡(2x)+15sin⁡(4x)+5sin⁡(8x)863 x + \frac{4 \sin^{5}{\left(2 x \right)}}{5} - 16 \sin^{3}{\left(2 x \right)} + 64 \sin{\left(2 x \right)} + 15 \sin{\left(4 x \right)} + \frac{5 \sin{\left(8 x \right)}}{8}

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

      ∫(−640cos⁡8(x)) dx=−640∫cos⁡8(x) dx\int \left(- 640 \cos^{8}{\left(x \right)}\right)\, dx = - 640 \int \cos^{8}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

        (cos⁡(2x)2+12)4=cos⁡4(2x)16+cos⁡3(2x)4+3cos⁡2(2x)8+cos⁡(2x)4+116\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{4} = \frac{\cos^{4}{\left(2 x \right)}}{16} + \frac{\cos^{3}{\left(2 x \right)}}{4} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} + \frac{\cos{\left(2 x \right)}}{4} + \frac{1}{16}

      3. Integrate term-by-term:

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

          ∫cos⁡4(2x)16 dx=∫cos⁡4(2x) dx16\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}

          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: 3x128+sin⁡(4x)128+sin⁡(8x)1024\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}

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

          ∫cos⁡3(2x)4 dx=∫cos⁡3(2x) dx4\int \frac{\cos^{3}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{4}

          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: −sin⁡3(2x)24+sin⁡(2x)8- \frac{\sin^{3}{\left(2 x \right)}}{24} + \frac{\sin{\left(2 x \right)}}{8}

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

          ∫3cos⁡2(2x)8 dx=3∫cos⁡2(2x) dx8\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}

          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: 3x16+3sin⁡(4x)64\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}

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

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

          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: sin⁡(2x)8\frac{\sin{\left(2 x \right)}}{8}

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

          ∫116 dx=x16\int \frac{1}{16}\, dx = \frac{x}{16}

        The result is: 35x128−sin⁡3(2x)24+sin⁡(2x)4+7sin⁡(4x)128+sin⁡(8x)1024\frac{35 x}{128} - \frac{\sin^{3}{\left(2 x \right)}}{24} + \frac{\sin{\left(2 x \right)}}{4} + \frac{7 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}

      So, the result is: −175x+80sin⁡3(2x)3−160sin⁡(2x)−35sin⁡(4x)−5sin⁡(8x)8- 175 x + \frac{80 \sin^{3}{\left(2 x \right)}}{3} - 160 \sin{\left(2 x \right)} - 35 \sin{\left(4 x \right)} - \frac{5 \sin{\left(8 x \right)}}{8}

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

      ∫560cos⁡6(x) dx=560∫cos⁡6(x) dx\int 560 \cos^{6}{\left(x \right)}\, dx = 560 \int \cos^{6}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

        (cos⁡(2x)2+12)3=cos⁡3(2x)8+3cos⁡2(2x)8+3cos⁡(2x)8+18\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{3} = \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} + \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}

      3. Integrate term-by-term:

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

          ∫cos⁡3(2x)8 dx=∫cos⁡3(2x) dx8\int \frac{\cos^{3}{\left(2 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}

          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: −sin⁡3(2x)48+sin⁡(2x)16- \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{16}

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

          ∫3cos⁡2(2x)8 dx=3∫cos⁡2(2x) dx8\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}

          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: 3x16+3sin⁡(4x)64\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}

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

          ∫3cos⁡(2x)8 dx=3∫cos⁡(2x) dx8\int \frac{3 \cos{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}

          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: 3sin⁡(2x)16\frac{3 \sin{\left(2 x \right)}}{16}

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

          ∫18 dx=x8\int \frac{1}{8}\, dx = \frac{x}{8}

        The result is: 5x16−sin⁡3(2x)48+sin⁡(2x)4+3sin⁡(4x)64\frac{5 x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}

      So, the result is: 175x−35sin⁡3(2x)3+140sin⁡(2x)+105sin⁡(4x)4175 x - \frac{35 \sin^{3}{\left(2 x \right)}}{3} + 140 \sin{\left(2 x \right)} + \frac{105 \sin{\left(4 x \right)}}{4}

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

      ∫(−196cos⁡4(x)) dx=−196∫cos⁡4(x) dx\int \left(- 196 \cos^{4}{\left(x \right)}\right)\, dx = - 196 \int \cos^{4}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

        (cos⁡(2x)2+12)2=cos⁡2(2x)4+cos⁡(2x)2+14\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} + \frac{\cos{\left(2 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(2x)4 dx=∫cos⁡2(2x) dx4\int \frac{\cos^{2}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}

          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: x8+sin⁡(4x)32\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}

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

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

          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: sin⁡(2x)4\frac{\sin{\left(2 x \right)}}{4}

        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⁡(2x)4+sin⁡(4x)32\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}

      So, the result is: −147x2−49sin⁡(2x)−49sin⁡(4x)8- \frac{147 x}{2} - 49 \sin{\left(2 x \right)} - \frac{49 \sin{\left(4 x \right)}}{8}

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

      ∫21cos⁡2(x) dx=21∫cos⁡2(x) dx\int 21 \cos^{2}{\left(x \right)}\, dx = 21 \int \cos^{2}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

          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: sin⁡(2x)4\frac{\sin{\left(2 x \right)}}{4}

        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⁡(2x)4\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}

      So, the result is: 21x2+21sin⁡(2x)4\frac{21 x}{2} + \frac{21 \sin{\left(2 x \right)}}{4}

    The result is: 4sin⁡5(2x)5−sin⁡3(2x)+sin⁡(2x)4+sin⁡(4x)8\frac{4 \sin^{5}{\left(2 x \right)}}{5} - \sin^{3}{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{8}

  3. Add the constant of integration:

    4sin⁡5(2x)5−sin⁡3(2x)+sin⁡(2x)4+sin⁡(4x)8+constant\frac{4 \sin^{5}{\left(2 x \right)}}{5} - \sin^{3}{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{8}+ \mathrm{constant}

The answer is:

4sin⁡5(2x)5−sin⁡3(2x)+sin⁡(2x)4+sin⁡(4x)8+constant\frac{4 \sin^{5}{\left(2 x \right)}}{5} - \sin^{3}{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{8}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                  5     
 |                               3        sin(2*x)   sin(4*x)   4*sin (2*x)
 | cos(7*x)*cos(3*x) dx = C - sin (2*x) + -------- + -------- + -----------
 |                                           4          8            5     
/
sin⁡(10 x)20+sin⁡(4 x)8{{\sin \left(10\,x\right)}\over{20}}+{{\sin \left(4\,x\right) }\over{8}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
Plot the graph f(x)
The answer [src] 
  3*cos(7)*sin(3)   7*cos(3)*sin(7)
- --------------- + ---------------
         40                40
2 sin⁡10+5 sin⁡440{{2\,\sin 10+5\,\sin 4}\over{40}} 
=
= 
  3*cos(7)*sin(3)   7*cos(3)*sin(7)
- --------------- + ---------------
         40                40
7sin⁡(7)cos⁡(3)40−3sin⁡(3)cos⁡(7)40\frac{7 \sin{\left(7 \right)} \cos{\left(3 \right)}}{40} - \frac{3 \sin{\left(3 \right)} \cos{\left(7 \right)}}{40}
Simplify
Numerical answer [src] 
-0.12180136745796
-0.12180136745796
The graph
Integral of cos(7x)cos(3x) dx

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

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