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sin^4(3x)
Solving integrals/ sin^4(3x)

Integral of sin^4(3x) dx

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∫01sin⁡4(3x) dx\int\limits_{0}^{1} \sin^{4}{\left(3 x \right)}\, dx 
Integral(sin(3*x)^4, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡4(3x)=(12−cos⁡(6x)2)2\sin^{4}{\left(3 x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2}

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (12−cos⁡(6x)2)2=cos⁡2(6x)4−cos⁡(6x)2+14\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(6 x \right)}}{4} - \frac{\cos{\left(6 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(6x)4 dx=∫cos⁡2(6x) dx4\int \frac{\cos^{2}{\left(6 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{4}

        1. Rewrite the integrand:

          cos⁡2(6x)=cos⁡(12x)2+12\cos^{2}{\left(6 x \right)} = \frac{\cos{\left(12 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⁡(12x)2 dx=∫cos⁡(12x) dx2\int \frac{\cos{\left(12 x \right)}}{2}\, dx = \frac{\int \cos{\left(12 x \right)}\, dx}{2}

            1. Let u=12xu = 12 x.

              Then let du=12dxdu = 12 dx and substitute du12\frac{du}{12}:

              ∫cos⁡(u)12 du\int \frac{\cos{\left(u \right)}}{12}\, du

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

                ∫cos⁡(u) du=∫cos⁡(u) du12\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}

                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)12\frac{\sin{\left(u \right)}}{12}

              Now substitute uu back in:

              sin⁡(12x)12\frac{\sin{\left(12 x \right)}}{12}

            So, the result is: sin⁡(12x)24\frac{\sin{\left(12 x \right)}}{24}

          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⁡(12x)24\frac{x}{2} + \frac{\sin{\left(12 x \right)}}{24}

        So, the result is: x8+sin⁡(12x)96\frac{x}{8} + \frac{\sin{\left(12 x \right)}}{96}

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

        ∫(−cos⁡(6x)2) dx=−∫cos⁡(6x) dx2\int \left(- \frac{\cos{\left(6 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{2}

        1. Let u=6xu = 6 x.

          Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

          ∫cos⁡(u)6 du\int \frac{\cos{\left(u \right)}}{6}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du6\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

            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)6\frac{\sin{\left(u \right)}}{6}

          Now substitute uu back in:

          sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

        So, the result is: −sin⁡(6x)12- \frac{\sin{\left(6 x \right)}}{12}

      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⁡(6x)12+sin⁡(12x)96\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}

    Method #2

    1. Rewrite the integrand:

      (12−cos⁡(6x)2)2=cos⁡2(6x)4−cos⁡(6x)2+14\left(\frac{1}{2} - \frac{\cos{\left(6 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(6 x \right)}}{4} - \frac{\cos{\left(6 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(6x)4 dx=∫cos⁡2(6x) dx4\int \frac{\cos^{2}{\left(6 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(6 x \right)}\, dx}{4}

        1. Rewrite the integrand:

          cos⁡2(6x)=cos⁡(12x)2+12\cos^{2}{\left(6 x \right)} = \frac{\cos{\left(12 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⁡(12x)2 dx=∫cos⁡(12x) dx2\int \frac{\cos{\left(12 x \right)}}{2}\, dx = \frac{\int \cos{\left(12 x \right)}\, dx}{2}

            1. Let u=12xu = 12 x.

              Then let du=12dxdu = 12 dx and substitute du12\frac{du}{12}:

              ∫cos⁡(u)12 du\int \frac{\cos{\left(u \right)}}{12}\, du

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

                ∫cos⁡(u) du=∫cos⁡(u) du12\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{12}

                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)12\frac{\sin{\left(u \right)}}{12}

              Now substitute uu back in:

              sin⁡(12x)12\frac{\sin{\left(12 x \right)}}{12}

            So, the result is: sin⁡(12x)24\frac{\sin{\left(12 x \right)}}{24}

          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⁡(12x)24\frac{x}{2} + \frac{\sin{\left(12 x \right)}}{24}

        So, the result is: x8+sin⁡(12x)96\frac{x}{8} + \frac{\sin{\left(12 x \right)}}{96}

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

        ∫(−cos⁡(6x)2) dx=−∫cos⁡(6x) dx2\int \left(- \frac{\cos{\left(6 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{2}

        1. Let u=6xu = 6 x.

          Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

          ∫cos⁡(u)6 du\int \frac{\cos{\left(u \right)}}{6}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du6\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

            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)6\frac{\sin{\left(u \right)}}{6}

          Now substitute uu back in:

          sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

        So, the result is: −sin⁡(6x)12- \frac{\sin{\left(6 x \right)}}{12}

      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⁡(6x)12+sin⁡(12x)96\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}

  3. Add the constant of integration:

    3x8−sin⁡(6x)12+sin⁡(12x)96+constant\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}+ \mathrm{constant}

The answer is:

3x8−sin⁡(6x)12+sin⁡(12x)96+constant\frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                             
 |                                              
 |    4               sin(6*x)   sin(12*x)   3*x
 | sin (3*x) dx = C - -------- + --------- + ---
 |                       12          96       8 
/
∫sin⁡4(3x) dx=C+3x8−sin⁡(6x)12+sin⁡(12x)96\int \sin^{4}{\left(3 x \right)}\, dx = C + \frac{3 x}{8} - \frac{\sin{\left(6 x \right)}}{12} + \frac{\sin{\left(12 x \right)}}{96}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9002
Plot the graph f(x)
The answer [src] 
                       3          
3   cos(3)*sin(3)   sin (3)*cos(3)
- - ------------- - --------------
8         8               12
−sin⁡3(3)cos⁡(3)12−sin⁡(3)cos⁡(3)8+38- \frac{\sin^{3}{\left(3 \right)} \cos{\left(3 \right)}}{12} - \frac{\sin{\left(3 \right)} \cos{\left(3 \right)}}{8} + \frac{3}{8} 
=
= 
                       3          
3   cos(3)*sin(3)   sin (3)*cos(3)
- - ------------- - --------------
8         8               12
−sin⁡3(3)cos⁡(3)12−sin⁡(3)cos⁡(3)8+38- \frac{\sin^{3}{\left(3 \right)} \cos{\left(3 \right)}}{12} - \frac{\sin{\left(3 \right)} \cos{\left(3 \right)}}{8} + \frac{3}{8} 
3/8 - cos(3)*sin(3)/8 - sin(3)^3*cos(3)/12
Numerical answer [src] 
0.392695323620739
0.392695323620739
The graph
Integral of sin^4(3x) dx

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

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