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cos^3(4x)

Integral of cos^3(4x) dx

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The solution

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01cos3(4x)dx\int\limits_{0}^{1} \cos^{3}{\left(4 x \right)}\, dx
Detail solution
  1. Rewrite the integrand:

    cos3(4x)=(1sin2(4x))cos(4x)\cos^{3}{\left(4 x \right)} = \left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)}

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=sin(4x)u = \sin{\left(4 x \right)}.

      Then let du=4cos(4x)dxdu = 4 \cos{\left(4 x \right)} dx and substitute dudu:

      (14u24)du\int \left(\frac{1}{4} - \frac{u^{2}}{4}\right)\, du

      1. Integrate term-by-term:

        1. The integral of a constant is the constant times the variable of integration:

          14du=u4\int \frac{1}{4}\, du = \frac{u}{4}

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

          (u24)du=u2du4\int \left(- \frac{u^{2}}{4}\right)\, du = - \frac{\int u^{2}\, du}{4}

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: u312- \frac{u^{3}}{12}

        The result is: u312+u4- \frac{u^{3}}{12} + \frac{u}{4}

      Now substitute uu back in:

      sin3(4x)12+sin(4x)4- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}

    Method #2

    1. Rewrite the integrand:

      (1sin2(4x))cos(4x)=sin2(4x)cos(4x)+cos(4x)\left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)} = - \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)} + \cos{\left(4 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:

        (sin2(4x)cos(4x))dx=sin2(4x)cos(4x)dx\int \left(- \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\right)\, dx = - \int \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\, dx

        1. Let u=sin(4x)u = \sin{\left(4 x \right)}.

          Then let du=4cos(4x)dxdu = 4 \cos{\left(4 x \right)} dx and substitute du4\frac{du}{4}:

          u216du\int \frac{u^{2}}{16}\, du

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

            u24du=u2du4\int \frac{u^{2}}{4}\, du = \frac{\int u^{2}\, du}{4}

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            So, the result is: u312\frac{u^{3}}{12}

          Now substitute uu back in:

          sin3(4x)12\frac{\sin^{3}{\left(4 x \right)}}{12}

        So, the result is: sin3(4x)12- \frac{\sin^{3}{\left(4 x \right)}}{12}

      1. Let u=4xu = 4 x.

        Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

        cos(u)16du\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)4du=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}

      The result is: sin3(4x)12+sin(4x)4- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}

    Method #3

    1. Rewrite the integrand:

      (1sin2(4x))cos(4x)=sin2(4x)cos(4x)+cos(4x)\left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)} = - \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)} + \cos{\left(4 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:

        (sin2(4x)cos(4x))dx=sin2(4x)cos(4x)dx\int \left(- \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\right)\, dx = - \int \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\, dx

        1. Let u=sin(4x)u = \sin{\left(4 x \right)}.

          Then let du=4cos(4x)dxdu = 4 \cos{\left(4 x \right)} dx and substitute du4\frac{du}{4}:

          u216du\int \frac{u^{2}}{16}\, du

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

            u24du=u2du4\int \frac{u^{2}}{4}\, du = \frac{\int u^{2}\, du}{4}

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            So, the result is: u312\frac{u^{3}}{12}

          Now substitute uu back in:

          sin3(4x)12\frac{\sin^{3}{\left(4 x \right)}}{12}

        So, the result is: sin3(4x)12- \frac{\sin^{3}{\left(4 x \right)}}{12}

      1. Let u=4xu = 4 x.

        Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

        cos(u)16du\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)4du=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}

      The result is: sin3(4x)12+sin(4x)4- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}

  3. Now simplify:

    (3sin2(4x))sin(4x)12\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}

  4. Add the constant of integration:

    (3sin2(4x))sin(4x)12+constant\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}+ \mathrm{constant}


The answer is:

(3sin2(4x))sin(4x)12+constant\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                       
 |                       3                
 |    3               sin (4*x)   sin(4*x)
 | cos (4*x) dx = C - --------- + --------
 |                        12         4    
/                                         
sin(4x)sin3(4x)34{{\sin \left(4\,x\right)-{{\sin ^3\left(4\,x\right)}\over{3}} }\over{4}}
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
The answer [src]
     3            
  sin (4)   sin(4)
- ------- + ------
     12       4   
sin343sin412-{{\sin ^34-3\,\sin 4}\over{12}}
=
=
     3            
  sin (4)   sin(4)
- ------- + ------
     12       4   
sin(4)4sin3(4)12\frac{\sin{\left(4 \right)}}{4} - \frac{\sin^{3}{\left(4 \right)}}{12}
Numerical answer [src]
-0.153079070328579
-0.153079070328579
The graph
Integral of cos^3(4x) dx

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