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sin^4

Integral of sin^4 dx

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01sin4(x)dx\int\limits_{0}^{1} \sin^{4}{\left(x \right)}\, dx
Integral(sin(x)^4, (x, 0, 1))
Detail solution
  1. Rewrite the integrand:

    sin4(x)=(12cos(2x)2)2\sin^{4}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2}

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (12cos(2x)2)2=cos2(2x)4cos(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}

    2. Integrate term-by-term:

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

        cos2(2x)4dx=cos2(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:

          cos2(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)2dx=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)4du\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:

            12dx=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)2du\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:

        14dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

      The result is: 3x8sin(2x)4+sin(4x)32\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}

    Method #2

    1. Rewrite the integrand:

      (12cos(2x)2)2=cos2(2x)4cos(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}

    2. Integrate term-by-term:

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

        cos2(2x)4dx=cos2(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:

          cos2(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)2dx=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)4du\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:

            12dx=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)2du\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:

        14dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

      The result is: 3x8sin(2x)4+sin(4x)32\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}

  3. Add the constant of integration:

    3x8sin(2x)4+sin(4x)32+constant\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}+ \mathrm{constant}


The answer is:

3x8sin(2x)4+sin(4x)32+constant\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                          
 |                                           
 |    4             sin(2*x)   sin(4*x)   3*x
 | sin (x) dx = C - -------- + -------- + ---
 |                     4          32       8 
/                                            
sin4(x)dx=C+3x8sin(2x)4+sin(4x)32\int \sin^{4}{\left(x \right)}\, dx = C + \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.01.0
The answer [src]
                         3          
3   3*cos(1)*sin(1)   sin (1)*cos(1)
- - --------------- - --------------
8          8                4       
3sin(1)cos(1)8sin3(1)cos(1)4+38- \frac{3 \sin{\left(1 \right)} \cos{\left(1 \right)}}{8} - \frac{\sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{4} + \frac{3}{8}
=
=
                         3          
3   3*cos(1)*sin(1)   sin (1)*cos(1)
- - --------------- - --------------
8          8                4       
3sin(1)cos(1)8sin3(1)cos(1)4+38- \frac{3 \sin{\left(1 \right)} \cos{\left(1 \right)}}{8} - \frac{\sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{4} + \frac{3}{8}
3/8 - 3*cos(1)*sin(1)/8 - sin(1)^3*cos(1)/4
Numerical answer [src]
0.124025565315207
0.124025565315207
The graph
Integral of sin^4 dx

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