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cos^3(2x)
Solving integrals/ cos^3(2x)

Integral of cos^3(2x) dx

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∫01cos⁡3(2x) dx\int\limits_{0}^{1} \cos^{3}{\left(2 x \right)}\, dx 
Integral(cos(2*x)^3, (x, 0, 1))
Detail solution

  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. There are multiple ways to do this integral.

    Method #1

    1. 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}

    Method #2

    1. Rewrite the integrand:

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

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

        1. 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 du2\frac{du}{2}:

          ∫u22 du\int \frac{u^{2}}{2}\, du

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

            ∫u2 du=∫u2 du2\int u^{2}\, 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}

          Now substitute uu back in:

          sin⁡3(2x)6\frac{\sin^{3}{\left(2 x \right)}}{6}

        So, the result is: −sin⁡3(2x)6- \frac{\sin^{3}{\left(2 x \right)}}{6}

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫cos⁡(u)2 du\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}

      The result is: −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

    Method #3

    1. Rewrite the integrand:

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

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

        1. 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 du2\frac{du}{2}:

          ∫u22 du\int \frac{u^{2}}{2}\, du

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

            ∫u2 du=∫u2 du2\int u^{2}\, 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}

          Now substitute uu back in:

          sin⁡3(2x)6\frac{\sin^{3}{\left(2 x \right)}}{6}

        So, the result is: −sin⁡3(2x)6- \frac{\sin^{3}{\left(2 x \right)}}{6}

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫cos⁡(u)2 du\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}

      The result is: −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

  3. Now simplify:

    (3−sin⁡2(2x))sin⁡(2x)6\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}

  4. Add the constant of integration:

    (3−sin⁡2(2x))sin⁡(2x)6+constant\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}+ \mathrm{constant}

The answer is:

(3−sin⁡2(2x))sin⁡(2x)6+constant\frac{\left(3 - \sin^{2}{\left(2 x \right)}\right) \sin{\left(2 x \right)}}{6}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                       
 |                                  3     
 |    3               sin(2*x)   sin (2*x)
 | cos (2*x) dx = C + -------- - ---------
 |                       2           6    
/
∫cos⁡3(2x) dx=C−sin⁡3(2x)6+sin⁡(2x)2\int \cos^{3}{\left(2 x \right)}\, dx = C - \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-1
Plot the graph f(x)
The answer [src] 
            3   
sin(2)   sin (2)
------ - -------
  2         6
−sin⁡3(2)6+sin⁡(2)2- \frac{\sin^{3}{\left(2 \right)}}{6} + \frac{\sin{\left(2 \right)}}{2} 
=
= 
            3   
sin(2)   sin (2)
------ - -------
  2         6
−sin⁡3(2)6+sin⁡(2)2- \frac{\sin^{3}{\left(2 \right)}}{6} + \frac{\sin{\left(2 \right)}}{2} 
sin(2)/2 - sin(2)^3/6
Numerical answer [src] 
0.329344222634675
0.329344222634675
The graph
Integral of cos^3(2x) dx

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

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