Mister Exam

Other calculators

sin^6xcos^4x
Solving integrals/ sin^6xcos^4x

Integral of sin^6xcos^4x dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1                   
  /                   
 |                    
 |     6       4      
 |  sin (x)*cos (x) dx
 |                    
/                     
0
∫01sin⁡6(x)cos⁡4(x) dx\int\limits_{0}^{1} \sin^{6}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx 
Integral(sin(x)^6*cos(x)^4, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡6(x)cos⁡4(x)=(12−cos⁡(2x)2)3(cos⁡(2x)2+12)2\sin^{6}{\left(x \right)} \cos^{4}{\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}

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

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

        1. Rewrite the integrand:

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

        2. Let u=2xu = 2 x.

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

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

          1. Integrate term-by-term:

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

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

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

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

                ∫u4 du\int u^{4}\, du

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

                  ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

                Now substitute uu back in:

                sin⁡5(u)5\frac{\sin^{5}{\left(u \right)}}{5}

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

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

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

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

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

                ∫u2 du\int u^{2}\, du

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

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

                Now substitute uu back in:

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

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

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

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

              1. The integral of cosine is sine:

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

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

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

          Now substitute uu back in:

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

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

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

        ∫cos⁡4(2x)32 dx=∫cos⁡4(2x) dx32\int \frac{\cos^{4}{\left(2 x \right)}}{32}\, dx = \frac{\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. There are multiple ways to do this integral.

          Method #1

          1. 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}

          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(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}

          Method #2

          1. 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}

          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(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: 3x256+sin⁡(4x)256+sin⁡(8x)2048\frac{3 x}{256} + \frac{\sin{\left(4 x \right)}}{256} + \frac{\sin{\left(8 x \right)}}{2048}

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

        ∫cos⁡3(2x)16 dx=∫cos⁡3(2x) dx16\int \frac{\cos^{3}{\left(2 x \right)}}{16}\, dx = \frac{\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: −sin⁡3(2x)96+sin⁡(2x)32- \frac{\sin^{3}{\left(2 x \right)}}{96} + \frac{\sin{\left(2 x \right)}}{32}

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

        ∫(−cos⁡2(2x)16) dx=−∫cos⁡2(2x) dx16\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{16}\right)\, dx = - \frac{\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: −x32−sin⁡(4x)128- \frac{x}{32} - \frac{\sin{\left(4 x \right)}}{128}

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

        ∫(−cos⁡(2x)32) dx=−∫cos⁡(2x) dx32\int \left(- \frac{\cos{\left(2 x \right)}}{32}\right)\, dx = - \frac{\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: −sin⁡(2x)64- \frac{\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: 3x256−sin⁡5(2x)320−sin⁡(4x)256+sin⁡(8x)2048\frac{3 x}{256} - \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin{\left(4 x \right)}}{256} + \frac{\sin{\left(8 x \right)}}{2048}

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

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

        1. Rewrite the integrand:

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

        2. Let u=2xu = 2 x.

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

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

          1. Integrate term-by-term:

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

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

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

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

                ∫u4 du\int u^{4}\, du

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

                  ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

                Now substitute uu back in:

                sin⁡5(u)5\frac{\sin^{5}{\left(u \right)}}{5}

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

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

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

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

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

                ∫u2 du\int u^{2}\, du

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

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

                Now substitute uu back in:

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

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

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

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

              1. The integral of cosine is sine:

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

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

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

          Now substitute uu back in:

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

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

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

        ∫cos⁡4(2x)32 dx=∫cos⁡4(2x) dx32\int \frac{\cos^{4}{\left(2 x \right)}}{32}\, dx = \frac{\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: 3x256+sin⁡(4x)256+sin⁡(8x)2048\frac{3 x}{256} + \frac{\sin{\left(4 x \right)}}{256} + \frac{\sin{\left(8 x \right)}}{2048}

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

        ∫cos⁡3(2x)16 dx=∫cos⁡3(2x) dx16\int \frac{\cos^{3}{\left(2 x \right)}}{16}\, dx = \frac{\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: −sin⁡3(2x)96+sin⁡(2x)32- \frac{\sin^{3}{\left(2 x \right)}}{96} + \frac{\sin{\left(2 x \right)}}{32}

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

        ∫(−cos⁡2(2x)16) dx=−∫cos⁡2(2x) dx16\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{16}\right)\, dx = - \frac{\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: −x32−sin⁡(4x)128- \frac{x}{32} - \frac{\sin{\left(4 x \right)}}{128}

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

        ∫(−cos⁡(2x)32) dx=−∫cos⁡(2x) dx32\int \left(- \frac{\cos{\left(2 x \right)}}{32}\right)\, dx = - \frac{\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: −sin⁡(2x)64- \frac{\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: 3x256−sin⁡5(2x)320−sin⁡(4x)256+sin⁡(8x)2048\frac{3 x}{256} - \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin{\left(4 x \right)}}{256} + \frac{\sin{\left(8 x \right)}}{2048}

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                                              
 |                                        5                      
 |    6       4             sin(4*x)   sin (2*x)   sin(8*x)   3*x
 | sin (x)*cos (x) dx = C - -------- - --------- + -------- + ---
 |                            256         320        2048     256
/
sin⁡(8 x)2+4 x8−sin⁡(4 x)2+x4−sin⁡5(2 x)1032{{{{{{{{\sin \left(8\,x\right)}\over{2}}+4\,x}\over{8}}-{{\sin \left(4\,x\right)}\over{2}}+x}\over{4}}-{{\sin ^5\left(2\,x\right) }\over{10}}}\over{32}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.000.05
Plot the graph f(x)
The answer [src] 
                           9                3                5                   7          
 3    3*cos(1)*sin(1)   sin (1)*cos(1)   sin (1)*cos(1)   sin (1)*cos(1)   11*sin (1)*cos(1)
--- - --------------- - -------------- - -------------- - -------------- + -----------------
256         256               10              128              160                 80
5 sin⁡8−40 sin⁡4−32 sin⁡52+12010240{{5\,\sin 8-40\,\sin 4-32\,\sin ^52+120}\over{10240}} 
=
= 
                           9                3                5                   7          
 3    3*cos(1)*sin(1)   sin (1)*cos(1)   sin (1)*cos(1)   sin (1)*cos(1)   11*sin (1)*cos(1)
--- - --------------- - -------------- - -------------- - -------------- + -----------------
256         256               10              128              160                 80
−sin⁡9(1)cos⁡(1)10−3sin⁡(1)cos⁡(1)256−sin⁡3(1)cos⁡(1)128−sin⁡5(1)cos⁡(1)160+3256+11sin⁡7(1)cos⁡(1)80- \frac{\sin^{9}{\left(1 \right)} \cos{\left(1 \right)}}{10} - \frac{3 \sin{\left(1 \right)} \cos{\left(1 \right)}}{256} - \frac{\sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{128} - \frac{\sin^{5}{\left(1 \right)} \cos{\left(1 \right)}}{160} + \frac{3}{256} + \frac{11 \sin^{7}{\left(1 \right)} \cos{\left(1 \right)}}{80}
Simplify
Numerical answer [src] 
0.0132155107176474
0.0132155107176474
The graph
Integral of sin^6xcos^4x dx

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

    © Mister Exam – Calculator