Mister Exam

Other calculators

sin^4xcos^4xdx
Solving integrals/ sin^4xcos^4xdx

Integral of sin^4xcos^4xdx dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

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

  1. Rewrite the integrand:

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

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

          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)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⁡2(2x)8) dx=−∫cos⁡2(2x) dx8\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\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: −x16−sin⁡(4x)64- \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64}

      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: 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}

    Method #2

    1. Rewrite the integrand:

      (12−cos⁡(2x)2)2(cos⁡(2x)2+12)2=cos⁡4(2x)16−cos⁡2(2x)8+116\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} = \frac{\cos^{4}{\left(2 x \right)}}{16} - \frac{\cos^{2}{\left(2 x \right)}}{8} + \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⁡2(2x)8) dx=−∫cos⁡2(2x) dx8\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\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: −x16−sin⁡(4x)64- \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64}

      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: 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}

  3. Add the constant of integration:

    3x128−sin⁡(4x)128+sin⁡(8x)1024+constant\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}+ \mathrm{constant}

The answer is:

3x128−sin⁡(4x)128+sin⁡(8x)1024+constant\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                  
 |                                                   
 |    4       4             sin(4*x)   sin(8*x)   3*x
 | sin (x)*cos (x) dx = C - -------- + -------- + ---
 |                            128        1024     128
/
∫sin⁡4(x)cos⁡4(x) dx=C+3x128−sin⁡(4x)128+sin⁡(8x)1024\int \sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = C + \frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.000.10
Plot the graph f(x)
The answer [src] 
                           3          
 3    3*cos(2)*sin(2)   sin (2)*cos(2)
--- - --------------- - --------------
128         256              128
−sin⁡3(2)cos⁡(2)128−3sin⁡(2)cos⁡(2)256+3128- \frac{\sin^{3}{\left(2 \right)} \cos{\left(2 \right)}}{128} - \frac{3 \sin{\left(2 \right)} \cos{\left(2 \right)}}{256} + \frac{3}{128} 
=
= 
                           3          
 3    3*cos(2)*sin(2)   sin (2)*cos(2)
--- - --------------- - --------------
128         256              128
−sin⁡3(2)cos⁡(2)128−3sin⁡(2)cos⁡(2)256+3128- \frac{\sin^{3}{\left(2 \right)} \cos{\left(2 \right)}}{128} - \frac{3 \sin{\left(2 \right)} \cos{\left(2 \right)}}{256} + \frac{3}{128} 
3/128 - 3*cos(2)*sin(2)/256 - sin(2)^3*cos(2)/128
Numerical answer [src] 
0.0303161896573113
0.0303161896573113
The graph
Integral of sin^4xcos^4xdx dx

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

    © Mister Exam – Calculator