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Solving integrals/ 2sin6xsin2xdx

Integral of 2sin6xsin2xdx dx

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∫01sin⁡(2x)2sin⁡(6x) dx\int\limits_{0}^{1} \sin{\left(2 x \right)} 2 \sin{\left(6 x \right)}\, dx 
Integral((2*sin(6*x))*sin(2*x), (x, 0, 1))
Detail solution

  1. 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⁡(u)sin⁡(3u) du\int \sin{\left(u \right)} \sin{\left(3 u \right)}\, du

      1. Rewrite the integrand:

        sin⁡(3u)sin⁡(u)=−4sin⁡4(u)+3sin⁡2(u)\sin{\left(3 u \right)} \sin{\left(u \right)} = - 4 \sin^{4}{\left(u \right)} + 3 \sin^{2}{\left(u \right)}

      2. Integrate term-by-term:

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

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

          1. Rewrite the integrand:

            sin⁡4(u)=(12−cos⁡(2u)2)2\sin^{4}{\left(u \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)^{2}

          2. There are multiple ways to do this integral.

            Method #1

            1. Rewrite the integrand:

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

            2. Integrate term-by-term:

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

                ∫cos⁡2(2u)4 du=∫cos⁡2(2u) du4\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}

                1. Rewrite the integrand:

                  cos⁡2(2u)=cos⁡(4u)2+12\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \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⁡(4u)2 du=∫cos⁡(4u) du2\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}

                    1. Let u=4uu = 4 u.

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

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

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

                  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}

                  The result is: u2+sin⁡(4u)8\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}

                So, the result is: u8+sin⁡(4u)32\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}

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

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

                1. Let u=2uu = 2 u.

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

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

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

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

                So, the result is: −sin⁡(2u)4- \frac{\sin{\left(2 u \right)}}{4}

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

                ∫14 du=u4\int \frac{1}{4}\, du = \frac{u}{4}

              The result is: 3u8−sin⁡(2u)4+sin⁡(4u)32\frac{3 u}{8} - \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}

            Method #2

            1. Rewrite the integrand:

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

            2. Integrate term-by-term:

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

                ∫cos⁡2(2u)4 du=∫cos⁡2(2u) du4\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}

                1. Rewrite the integrand:

                  cos⁡2(2u)=cos⁡(4u)2+12\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \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⁡(4u)2 du=∫cos⁡(4u) du2\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}

                    1. Let u=4uu = 4 u.

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

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

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

                  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}

                  The result is: u2+sin⁡(4u)8\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}

                So, the result is: u8+sin⁡(4u)32\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}

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

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

                1. Let u=2uu = 2 u.

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

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

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

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

                So, the result is: −sin⁡(2u)4- \frac{\sin{\left(2 u \right)}}{4}

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

                ∫14 du=u4\int \frac{1}{4}\, du = \frac{u}{4}

              The result is: 3u8−sin⁡(2u)4+sin⁡(4u)32\frac{3 u}{8} - \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}

          So, the result is: −3u2+sin⁡(2u)−sin⁡(4u)8- \frac{3 u}{2} + \sin{\left(2 u \right)} - \frac{\sin{\left(4 u \right)}}{8}

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

          ∫3sin⁡2(u) du=3∫sin⁡2(u) du\int 3 \sin^{2}{\left(u \right)}\, du = 3 \int \sin^{2}{\left(u \right)}\, du

          1. Rewrite the integrand:

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

          2. 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:

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

              1. Let u=2uu = 2 u.

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

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

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

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

              So, the result is: −sin⁡(2u)4- \frac{\sin{\left(2 u \right)}}{4}

            The result is: u2−sin⁡(2u)4\frac{u}{2} - \frac{\sin{\left(2 u \right)}}{4}

          So, the result is: 3u2−3sin⁡(2u)4\frac{3 u}{2} - \frac{3 \sin{\left(2 u \right)}}{4}

        The result is: sin⁡(2u)4−sin⁡(4u)8\frac{\sin{\left(2 u \right)}}{4} - \frac{\sin{\left(4 u \right)}}{8}

      Now substitute uu back in:

      sin⁡(4x)4−sin⁡(8x)8\frac{\sin{\left(4 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{8}

    Method #2

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

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

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

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

          ∫32sin⁡6(x)cos⁡2(x) dx=32∫sin⁡6(x)cos⁡2(x) dx\int 32 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 32 \int \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

          1. Rewrite the integrand:

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

          2. Let u=2xu = 2 x.

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

            ∫(−cos⁡4(u)32+cos⁡3(u)16−cos⁡(u)16+132) du\int \left(- \frac{\cos^{4}{\left(u \right)}}{32} + \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{32}\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:

                ∫(−cos⁡4(u)32) du=−∫cos⁡4(u) du32\int \left(- \frac{\cos^{4}{\left(u \right)}}{32}\right)\, du = - \frac{\int \cos^{4}{\left(u \right)}\, du}{32}

                1. Rewrite the integrand:

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

                2. Rewrite the integrand:

                  (cos⁡(2u)2+12)2=cos⁡2(2u)4+cos⁡(2u)2+14\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 u \right)}}{4} + \frac{\cos{\left(2 u \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(2u)4 du=∫cos⁡2(2u) du4\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}

                    1. Rewrite the integrand:

                      cos⁡2(2u)=cos⁡(4u)2+12\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \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⁡(4u)2 du=∫cos⁡(4u) du2\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}

                        1. Let u=4uu = 4 u.

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

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

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

                      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}

                      The result is: u2+sin⁡(4u)8\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}

                    So, the result is: u8+sin⁡(4u)32\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}

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

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

                    1. Let u=2uu = 2 u.

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

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

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

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

                    So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{4}

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

                    ∫14 du=u4\int \frac{1}{4}\, du = \frac{u}{4}

                  The result is: 3u8+sin⁡(2u)4+sin⁡(4u)32\frac{3 u}{8} + \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}

                So, the result is: −3u256−sin⁡(2u)128−sin⁡(4u)1024- \frac{3 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024}

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

                ∫cos⁡3(u)16 du=∫cos⁡3(u) du16\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}

                1. Rewrite the integrand:

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

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

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

                  ∫(1−u2) du\int \left(1 - u^{2}\right)\, du

                  1. Integrate term-by-term:

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

                      ∫1 du=u\int 1\, du = u

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

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

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

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

                      So, the result is: −u33- \frac{u^{3}}{3}

                    The result is: −u33+u- \frac{u^{3}}{3} + u

                  Now substitute uu back in:

                  −sin⁡3(u)3+sin⁡(u)- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}

                So, the result is: −sin⁡3(u)48+sin⁡(u)16- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}

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

                ∫(−cos⁡(u)16) du=−∫cos⁡(u) du16\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}

                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)16- \frac{\sin{\left(u \right)}}{16}

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

                ∫132 du=u32\int \frac{1}{32}\, du = \frac{u}{32}

              The result is: 5u256−sin⁡(2u)128−sin⁡(4u)1024−sin⁡3(u)48\frac{5 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024} - \frac{\sin^{3}{\left(u \right)}}{48}

            Now substitute uu back in:

            5x128−sin⁡3(2x)48−sin⁡(4x)128−sin⁡(8x)1024\frac{5 x}{128} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{128} - \frac{\sin{\left(8 x \right)}}{1024}

          So, the result is: 5x4−2sin⁡3(2x)3−sin⁡(4x)4−sin⁡(8x)32\frac{5 x}{4} - \frac{2 \sin^{3}{\left(2 x \right)}}{3} - \frac{\sin{\left(4 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{32}

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

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

          1. Rewrite the integrand:

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

          2. Let u=2xu = 2 x.

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

            ∫(cos⁡3(u)16−cos⁡2(u)16−cos⁡(u)16+116) du\int \left(\frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\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:

                ∫cos⁡3(u)16 du=∫cos⁡3(u) du16\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}

                1. Rewrite the integrand:

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

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

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

                  ∫(1−u2) du\int \left(1 - u^{2}\right)\, du

                  1. Integrate term-by-term:

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

                      ∫1 du=u\int 1\, du = u

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

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

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

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

                      So, the result is: −u33- \frac{u^{3}}{3}

                    The result is: −u33+u- \frac{u^{3}}{3} + u

                  Now substitute uu back in:

                  −sin⁡3(u)3+sin⁡(u)- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}

                So, the result is: −sin⁡3(u)48+sin⁡(u)16- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}

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

                ∫(−cos⁡2(u)16) du=−∫cos⁡2(u) du16\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}

                1. Rewrite the integrand:

                  cos⁡2(u)=cos⁡(2u)2+12\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \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⁡(2u)2 du=∫cos⁡(2u) du2\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}

                    1. Let u=2uu = 2 u.

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

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

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

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

                    So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{4}

                  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}

                  The result is: u2+sin⁡(2u)4\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}

                So, the result is: −u32−sin⁡(2u)64- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}

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

                ∫(−cos⁡(u)16) du=−∫cos⁡(u) du16\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}

                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)16- \frac{\sin{\left(u \right)}}{16}

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

                ∫116 du=u16\int \frac{1}{16}\, du = \frac{u}{16}

              The result is: u32−sin⁡(2u)64−sin⁡3(u)48\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} - \frac{\sin^{3}{\left(u \right)}}{48}

            Now substitute uu back in:

            x16−sin⁡3(2x)48−sin⁡(4x)64\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}

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

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

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

          1. Rewrite the integrand:

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

          2. Let u=2xu = 2 x.

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

            ∫(18−cos⁡2(u)8) du\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du

            1. Integrate term-by-term:

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

                ∫18 du=u8\int \frac{1}{8}\, du = \frac{u}{8}

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

                ∫(−cos⁡2(u)8) du=−∫cos⁡2(u) du8\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}

                1. Rewrite the integrand:

                  cos⁡2(u)=cos⁡(2u)2+12\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \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⁡(2u)2 du=∫cos⁡(2u) du2\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}

                    1. Let u=2uu = 2 u.

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

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

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

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

                    So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{4}

                  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}

                  The result is: u2+sin⁡(2u)4\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}

                So, the result is: −u16−sin⁡(2u)32- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}

              The result is: u16−sin⁡(2u)32\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}

            Now substitute uu back in:

            x8−sin⁡(4x)32\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}

          So, the result is: 3x4−3sin⁡(4x)16\frac{3 x}{4} - \frac{3 \sin{\left(4 x \right)}}{16}

        The result is: sin⁡(4x)16−sin⁡(8x)32\frac{\sin{\left(4 x \right)}}{16} - \frac{\sin{\left(8 x \right)}}{32}

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

    Method #3

    1. Rewrite the integrand:

      sin⁡(2x)2sin⁡(6x)=128sin⁡6(x)cos⁡2(x)−128sin⁡4(x)cos⁡2(x)+24sin⁡2(x)cos⁡2(x)\sin{\left(2 x \right)} 2 \sin{\left(6 x \right)} = 128 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)} - 128 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} + 24 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}

    2. Integrate term-by-term:

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

        ∫128sin⁡6(x)cos⁡2(x) dx=128∫sin⁡6(x)cos⁡2(x) dx\int 128 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 128 \int \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

        1. Rewrite the integrand:

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

        2. Let u=2xu = 2 x.

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

          ∫(−cos⁡4(u)32+cos⁡3(u)16−cos⁡(u)16+132) du\int \left(- \frac{\cos^{4}{\left(u \right)}}{32} + \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{32}\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:

              ∫(−cos⁡4(u)32) du=−∫cos⁡4(u) du32\int \left(- \frac{\cos^{4}{\left(u \right)}}{32}\right)\, du = - \frac{\int \cos^{4}{\left(u \right)}\, du}{32}

              1. Rewrite the integrand:

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

              2. Rewrite the integrand:

                (cos⁡(2u)2+12)2=cos⁡2(2u)4+cos⁡(2u)2+14\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 u \right)}}{4} + \frac{\cos{\left(2 u \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(2u)4 du=∫cos⁡2(2u) du4\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}

                  1. Rewrite the integrand:

                    cos⁡2(2u)=cos⁡(4u)2+12\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \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⁡(4u)2 du=∫cos⁡(4u) du2\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}

                      1. Let u=4uu = 4 u.

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

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

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

                    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}

                    The result is: u2+sin⁡(4u)8\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}

                  So, the result is: u8+sin⁡(4u)32\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}

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

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

                  1. Let u=2uu = 2 u.

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

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

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

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

                  So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{4}

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

                  ∫14 du=u4\int \frac{1}{4}\, du = \frac{u}{4}

                The result is: 3u8+sin⁡(2u)4+sin⁡(4u)32\frac{3 u}{8} + \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}

              So, the result is: −3u256−sin⁡(2u)128−sin⁡(4u)1024- \frac{3 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024}

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

              ∫cos⁡3(u)16 du=∫cos⁡3(u) du16\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}

              1. Rewrite the integrand:

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

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

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

                ∫(1−u2) du\int \left(1 - u^{2}\right)\, du

                1. Integrate term-by-term:

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

                    ∫1 du=u\int 1\, du = u

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

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

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

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

                    So, the result is: −u33- \frac{u^{3}}{3}

                  The result is: −u33+u- \frac{u^{3}}{3} + u

                Now substitute uu back in:

                −sin⁡3(u)3+sin⁡(u)- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}

              So, the result is: −sin⁡3(u)48+sin⁡(u)16- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}

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

              ∫(−cos⁡(u)16) du=−∫cos⁡(u) du16\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}

              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)16- \frac{\sin{\left(u \right)}}{16}

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

              ∫132 du=u32\int \frac{1}{32}\, du = \frac{u}{32}

            The result is: 5u256−sin⁡(2u)128−sin⁡(4u)1024−sin⁡3(u)48\frac{5 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024} - \frac{\sin^{3}{\left(u \right)}}{48}

          Now substitute uu back in:

          5x128−sin⁡3(2x)48−sin⁡(4x)128−sin⁡(8x)1024\frac{5 x}{128} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{128} - \frac{\sin{\left(8 x \right)}}{1024}

        So, the result is: 5x−8sin⁡3(2x)3−sin⁡(4x)−sin⁡(8x)85 x - \frac{8 \sin^{3}{\left(2 x \right)}}{3} - \sin{\left(4 x \right)} - \frac{\sin{\left(8 x \right)}}{8}

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

        ∫(−128sin⁡4(x)cos⁡2(x)) dx=−128∫sin⁡4(x)cos⁡2(x) dx\int \left(- 128 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 128 \int \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

        1. Rewrite the integrand:

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

        2. Let u=2xu = 2 x.

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

          ∫(cos⁡3(u)16−cos⁡2(u)16−cos⁡(u)16+116) du\int \left(\frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\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:

              ∫cos⁡3(u)16 du=∫cos⁡3(u) du16\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}

              1. Rewrite the integrand:

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

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

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

                ∫(1−u2) du\int \left(1 - u^{2}\right)\, du

                1. Integrate term-by-term:

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

                    ∫1 du=u\int 1\, du = u

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

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

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

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

                    So, the result is: −u33- \frac{u^{3}}{3}

                  The result is: −u33+u- \frac{u^{3}}{3} + u

                Now substitute uu back in:

                −sin⁡3(u)3+sin⁡(u)- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}

              So, the result is: −sin⁡3(u)48+sin⁡(u)16- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}

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

              ∫(−cos⁡2(u)16) du=−∫cos⁡2(u) du16\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}

              1. Rewrite the integrand:

                cos⁡2(u)=cos⁡(2u)2+12\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \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⁡(2u)2 du=∫cos⁡(2u) du2\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}

                  1. Let u=2uu = 2 u.

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

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

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

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

                  So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{4}

                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}

                The result is: u2+sin⁡(2u)4\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}

              So, the result is: −u32−sin⁡(2u)64- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}

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

              ∫(−cos⁡(u)16) du=−∫cos⁡(u) du16\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}

              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)16- \frac{\sin{\left(u \right)}}{16}

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

              ∫116 du=u16\int \frac{1}{16}\, du = \frac{u}{16}

            The result is: u32−sin⁡(2u)64−sin⁡3(u)48\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} - \frac{\sin^{3}{\left(u \right)}}{48}

          Now substitute uu back in:

          x16−sin⁡3(2x)48−sin⁡(4x)64\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}

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

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

        ∫24sin⁡2(x)cos⁡2(x) dx=24∫sin⁡2(x)cos⁡2(x) dx\int 24 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 24 \int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

        1. Rewrite the integrand:

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

        2. Let u=2xu = 2 x.

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

          ∫(18−cos⁡2(u)8) du\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du

          1. Integrate term-by-term:

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

              ∫18 du=u8\int \frac{1}{8}\, du = \frac{u}{8}

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

              ∫(−cos⁡2(u)8) du=−∫cos⁡2(u) du8\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}

              1. Rewrite the integrand:

                cos⁡2(u)=cos⁡(2u)2+12\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \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⁡(2u)2 du=∫cos⁡(2u) du2\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}

                  1. Let u=2uu = 2 u.

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

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

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

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

                  So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{4}

                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}

                The result is: u2+sin⁡(2u)4\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}

              So, the result is: −u16−sin⁡(2u)32- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}

            The result is: u16−sin⁡(2u)32\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}

          Now substitute uu back in:

          x8−sin⁡(4x)32\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}

        So, the result is: 3x−3sin⁡(4x)43 x - \frac{3 \sin{\left(4 x \right)}}{4}

      The result is: sin⁡(4x)4−sin⁡(8x)8\frac{\sin{\left(4 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{8}

  2. Add the constant of integration:

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

The answer is:

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

The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
  3*cos(6)*sin(2)   cos(2)*sin(6)
- --------------- + -------------
         8                8
−3sin⁡(2)cos⁡(6)8+sin⁡(6)cos⁡(2)8- \frac{3 \sin{\left(2 \right)} \cos{\left(6 \right)}}{8} + \frac{\sin{\left(6 \right)} \cos{\left(2 \right)}}{8} 
=
= 
  3*cos(6)*sin(2)   cos(2)*sin(6)
- --------------- + -------------
         8                8
−3sin⁡(2)cos⁡(6)8+sin⁡(6)cos⁡(2)8- \frac{3 \sin{\left(2 \right)} \cos{\left(6 \right)}}{8} + \frac{\sin{\left(6 \right)} \cos{\left(2 \right)}}{8} 
-3*cos(6)*sin(2)/8 + cos(2)*sin(6)/8
Numerical answer [src] 
-0.312870404654905
-0.312870404654905

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

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