Mister Exam

Other calculators

cos^5(2x)
Solving integrals/ cos^5(2x)

Integral of cos^5(2x) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1             
  /             
 |              
 |     5        
 |  cos (2*x) dx
 |              
/               
0
∫01cos⁡5(2x) dx\int\limits_{0}^{1} \cos^{5}{\left(2 x \right)}\, dx
Simplify
Detail solution

  1. Rewrite the integrand:

    cos⁡5(2x)=(1−sin⁡2(2x))2cos⁡(2x)\cos^{5}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right)^{2} \cos{\left(2 x \right)}

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=2xu = 2 x.

      Then let du=2dxdu = 2 dx and substitute dudu:

      ∫(sin⁡4(u)cos⁡(u)2−sin⁡2(u)cos⁡(u)+cos⁡(u)2) du\int \left(\frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2} - \sin^{2}{\left(u \right)} \cos{\left(u \right)} + \frac{\cos{\left(u \right)}}{2}\right)\, du

      1. Integrate term-by-term:

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

          ∫sin⁡4(u)cos⁡(u)2 du=∫sin⁡4(u)cos⁡(u) du2\int \frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2}\, du = \frac{\int \sin^{4}{\left(u \right)} \cos{\left(u \right)}\, du}{2}

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

            Then let du=cos⁡(u)dudu = \cos{\left(u \right)} du and substitute dudu:

            ∫u4 du\int u^{4}\, du

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

              ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

            Now substitute uu back in:

            sin⁡5(u)5\frac{\sin^{5}{\left(u \right)}}{5}

          So, the result is: sin⁡5(u)10\frac{\sin^{5}{\left(u \right)}}{10}

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

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

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

            Then let du=cos⁡(u)dudu = \cos{\left(u \right)} du and substitute dudu:

            ∫u2 du\int u^{2}\, du

            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}

            Now substitute uu back in:

            sin⁡3(u)3\frac{\sin^{3}{\left(u \right)}}{3}

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

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

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

        The result is: sin⁡5(u)10−sin⁡3(u)3+sin⁡(u)2\frac{\sin^{5}{\left(u \right)}}{10} - \frac{\sin^{3}{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{2}

      Now substitute uu back in:

      sin⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      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}:

        ∫u44 du\int \frac{u^{4}}{4}\, du

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

          ∫u42 du=∫u4 du2\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}

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

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          So, the result is: u510\frac{u^{5}}{10}

        Now substitute uu back in:

        sin⁡5(2x)10\frac{\sin^{5}{\left(2 x \right)}}{10}

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

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

          ∫u24 du\int \frac{u^{2}}{4}\, du

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

            ∫u22 du=∫u2 du2\int \frac{u^{2}}{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)3- \frac{\sin^{3}{\left(2 x \right)}}{3}

      1. Let u=2xu = 2 x.

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

        ∫cos⁡(u)4 du\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)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, 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⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}

    Method #3

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      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}:

        ∫u44 du\int \frac{u^{4}}{4}\, du

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

          ∫u42 du=∫u4 du2\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}

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

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          So, the result is: u510\frac{u^{5}}{10}

        Now substitute uu back in:

        sin⁡5(2x)10\frac{\sin^{5}{\left(2 x \right)}}{10}

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

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

          ∫u24 du\int \frac{u^{2}}{4}\, du

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

            ∫u22 du=∫u2 du2\int \frac{u^{2}}{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)3- \frac{\sin^{3}{\left(2 x \right)}}{3}

      1. Let u=2xu = 2 x.

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

        ∫cos⁡(u)4 du\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)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, 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⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}

  3. Add the constant of integration:

    sin⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2+constant\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}+ \mathrm{constant}

The answer is:

sin⁡5(2x)10−sin⁡3(2x)3+sin⁡(2x)2+constant\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                   
 |                                  3           5     
 |    5               sin(2*x)   sin (2*x)   sin (2*x)
 | cos (2*x) dx = C + -------- - --------- + ---------
 |                       2           3           10   
/
sin⁡5(2 x)5−2 sin⁡3(2 x)3+sin⁡(2 x)2{{{{\sin ^5\left(2\,x\right)}\over{5}}-{{2\,\sin ^3\left(2\,x \right)}\over{3}}+\sin \left(2\,x\right)}\over{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         5   
sin(2)   sin (2)   sin (2)
------ - ------- + -------
  2         3         10
3 sin⁡52−10 sin⁡32+15 sin⁡230{{3\,\sin ^52-10\,\sin ^32+15\,\sin 2}\over{30}} 
=
= 
            3         5   
sin(2)   sin (2)   sin (2)
------ - ------- + -------
  2         3         10
−sin⁡3(2)3+sin⁡5(2)10+sin⁡(2)2- \frac{\sin^{3}{\left(2 \right)}}{3} + \frac{\sin^{5}{\left(2 \right)}}{10} + \frac{\sin{\left(2 \right)}}{2}
Simplify
Numerical answer [src] 
0.266202423408773
0.266202423408773
The graph
Integral of cos^5(2x) dx

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

    © Mister Exam – Calculator