Mister Exam

Other calculators

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

Integral of sin^2(2x)cos^2(2x) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  p                       
  -                       
  2                       
  /                       
 |                        
 |     2         2        
 |  sin (2*x)*cos (2*x) dx
 |                        
/                         
0
∫0p2sin⁡2(2x)cos⁡2(2x) dx\int\limits_{0}^{\frac{p}{2}} \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)}\, dx 
Integral(sin(2*x)^2*cos(2*x)^2, (x, 0, p/2))
Detail solution

  1. Rewrite the integrand:

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

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=4xu = 4 x.

      Then let du=4dxdu = 4 dx and substitute dudu:

      ∫(116−cos⁡2(u)16) du\int \left(\frac{1}{16} - \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du

      1. Integrate term-by-term:

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

          ∫116 du=u16\int \frac{1}{16}\, du = \frac{u}{16}

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

          ∫(−cos⁡2(u)16) du=−∫cos⁡2(u) du16\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}

          1. Rewrite the integrand:

            cos⁡2(u)=cos⁡(2u)2+12\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \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⁡(2u)2 du=∫cos⁡(2u) du2\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}

              1. Let u=2uu = 2 u.

                Then let du=2dudu = 2 du 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⁡(2u)2\frac{\sin{\left(2 u \right)}}{2}

              So, the result is: sin⁡(2u)4\frac{\sin{\left(2 u \right)}}{4}

            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}

            The result is: u2+sin⁡(2u)4\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}

          So, the result is: −u32−sin⁡(2u)64- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}

        The result is: u32−sin⁡(2u)64\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}

      Now substitute uu back in:

      x8−sin⁡(8x)64\frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}

    Method #2

    1. Rewrite the integrand:

      (12−cos⁡(4x)2)(cos⁡(4x)2+12)=14−cos⁡2(4x)4\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right) \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(4 x \right)}}{4}

    2. Integrate term-by-term:

      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}

      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 \left(- \frac{\cos^{2}{\left(4 x \right)}}{4}\right)\, 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}

      The result is: x8−sin⁡(8x)64\frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}

    Method #3

    1. Rewrite the integrand:

      (12−cos⁡(4x)2)(cos⁡(4x)2+12)=14−cos⁡2(4x)4\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right) \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right) = \frac{1}{4} - \frac{\cos^{2}{\left(4 x \right)}}{4}

    2. Integrate term-by-term:

      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}

      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 \left(- \frac{\cos^{2}{\left(4 x \right)}}{4}\right)\, 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}

      The result is: x8−sin⁡(8x)64\frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}

  3. Add the constant of integration:

    x8−sin⁡(8x)64+constant\frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}

The answer is:

x8−sin⁡(8x)64+constant\frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                         
 |                                          
 |    2         2               sin(8*x)   x
 | sin (2*x)*cos (2*x) dx = C - -------- + -
 |                                 64      8
/
∫sin⁡2(2x)cos⁡2(2x) dx=C+x8−sin⁡(8x)64\int \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)}\, dx = C + \frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}
Simplify
The answer [src] 
p    cos(2*p)*sin(2*p)
-- - -----------------
16           32
p16−sin⁡(2p)cos⁡(2p)32\frac{p}{16} - \frac{\sin{\left(2 p \right)} \cos{\left(2 p \right)}}{32} 
=
= 
p    cos(2*p)*sin(2*p)
-- - -----------------
16           32
p16−sin⁡(2p)cos⁡(2p)32\frac{p}{16} - \frac{\sin{\left(2 p \right)} \cos{\left(2 p \right)}}{32} 
p/16 - cos(2*p)*sin(2*p)/32

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

    © Mister Exam – Calculator