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Solving integrals/ sin(5*x)*sin(3*x)

Integral of sin(5*x)*sin(3*x) dx

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∫01sin⁡(3x)sin⁡(5x) dx\int\limits_{0}^{1} \sin{\left(3 x \right)} \sin{\left(5 x \right)}\, dx 
Integral(sin(5*x)*sin(3*x), (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡(3x)sin⁡(5x)=−64sin⁡8(x)+128sin⁡6(x)−80sin⁡4(x)+15sin⁡2(x)\sin{\left(3 x \right)} \sin{\left(5 x \right)} = - 64 \sin^{8}{\left(x \right)} + 128 \sin^{6}{\left(x \right)} - 80 \sin^{4}{\left(x \right)} + 15 \sin^{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:

      ∫(−64sin⁡8(x)) dx=−64∫sin⁡8(x) dx\int \left(- 64 \sin^{8}{\left(x \right)}\right)\, dx = - 64 \int \sin^{8}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. There are multiple ways to do this integral.

        Method #1

        1. Rewrite the integrand:

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

        2. 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)8 du\int \frac{\cos{\left(u \right)}}{8}\, du

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

                        ∫cos⁡(u) du=∫cos⁡(u) du8\int \cos{\left(u \right)}\, 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)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⁡(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 \left(- \frac{\cos^{3}{\left(2 x \right)}}{4}\right)\, 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. There are multiple ways to do this integral.

              Method #1

              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 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}

              Method #2

              1. Rewrite the integrand:

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

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

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

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

                      ∫u2 du=∫u2 du2\int u^{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)6- \frac{\sin^{3}{\left(2 x \right)}}{6}

                1. Let u=2xu = 2 x.

                  Then let du=2dxdu = 2 dx 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⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

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

              Method #3

              1. Rewrite the integrand:

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

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

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

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

                      ∫u2 du=∫u2 du2\int u^{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)6- \frac{\sin^{3}{\left(2 x \right)}}{6}

                1. Let u=2xu = 2 x.

                  Then let du=2dxdu = 2 dx 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⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

                The result is: −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)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⁡(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 \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, 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)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⁡(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}

        Method #2

        1. Rewrite the integrand:

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

        2. 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)8 du\int \frac{\cos{\left(u \right)}}{8}\, du

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

                        ∫cos⁡(u) du=∫cos⁡(u) du8\int \cos{\left(u \right)}\, 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)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⁡(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 \left(- \frac{\cos^{3}{\left(2 x \right)}}{4}\right)\, 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)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⁡(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 \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, 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)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⁡(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: −35x2−8sin⁡3(2x)3+16sin⁡(2x)−7sin⁡(4x)2−sin⁡(8x)16- \frac{35 x}{2} - \frac{8 \sin^{3}{\left(2 x \right)}}{3} + 16 \sin{\left(2 x \right)} - \frac{7 \sin{\left(4 x \right)}}{2} - \frac{\sin{\left(8 x \right)}}{16}

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

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

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

        (12−cos⁡(2x)2)3=−cos⁡3(2x)8+3cos⁡2(2x)8−3cos⁡(2x)8+18\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{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 \left(- \frac{\cos^{3}{\left(2 x \right)}}{8}\right)\, 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)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⁡(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 \left(- \frac{3 \cos{\left(2 x \right)}}{8}\right)\, 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)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⁡(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: 40x+8sin⁡3(2x)3−32sin⁡(2x)+6sin⁡(4x)40 x + \frac{8 \sin^{3}{\left(2 x \right)}}{3} - 32 \sin{\left(2 x \right)} + 6 \sin{\left(4 x \right)}

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

      ∫(−80sin⁡4(x)) dx=−80∫sin⁡4(x) dx\int \left(- 80 \sin^{4}{\left(x \right)}\right)\, dx = - 80 \int \sin^{4}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

        (12−cos⁡(2x)2)2=cos⁡2(2x)4−cos⁡(2x)2+14\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{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)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⁡(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 \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, 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)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⁡(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: −30x+20sin⁡(2x)−5sin⁡(4x)2- 30 x + 20 \sin{\left(2 x \right)} - \frac{5 \sin{\left(4 x \right)}}{2}

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

      ∫15sin⁡2(x) dx=15∫sin⁡2(x) dx\int 15 \sin^{2}{\left(x \right)}\, dx = 15 \int \sin^{2}{\left(x \right)}\, dx

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

        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}

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

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

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

      So, the result is: 15x2−15sin⁡(2x)4\frac{15 x}{2} - \frac{15 \sin{\left(2 x \right)}}{4}

    The result is: sin⁡(2x)4−sin⁡(8x)16\frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{16}

  3. Add the constant of integration:

    sin⁡(2x)4−sin⁡(8x)16+constant\frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{16}+ \mathrm{constant}

The answer is:

sin⁡(2x)4−sin⁡(8x)16+constant\frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{16}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                              
 |                            sin(8*x)   sin(2*x)
 | sin(5*x)*sin(3*x) dx = C - -------- + --------
 |                               16         4    
/
∫sin⁡(3x)sin⁡(5x) dx=C+sin⁡(2x)4−sin⁡(8x)16\int \sin{\left(3 x \right)} \sin{\left(5 x \right)}\, dx = C + \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{16}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
Plot the graph f(x)
The answer [src] 
  5*cos(5)*sin(3)   3*cos(3)*sin(5)
- --------------- + ---------------
         16                16
−5sin⁡(3)cos⁡(5)16+3sin⁡(5)cos⁡(3)16- \frac{5 \sin{\left(3 \right)} \cos{\left(5 \right)}}{16} + \frac{3 \sin{\left(5 \right)} \cos{\left(3 \right)}}{16} 
=
= 
  5*cos(5)*sin(3)   3*cos(3)*sin(5)
- --------------- + ---------------
         16                16
−5sin⁡(3)cos⁡(5)16+3sin⁡(5)cos⁡(3)16- \frac{5 \sin{\left(3 \right)} \cos{\left(5 \right)}}{16} + \frac{3 \sin{\left(5 \right)} \cos{\left(3 \right)}}{16} 
-5*cos(5)*sin(3)/16 + 3*cos(3)*sin(5)/16
Numerical answer [src] 
0.165489466292459
0.165489466292459

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

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