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sin(2*x)/cos(2*x)

Integral of sin(2*x)/cos(2*x) dx

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The solution

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 |  sin(2*x)   
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 |  cos(2*x)   
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01sin(2x)cos(2x)dx\int\limits_{0}^{1} \frac{\sin{\left(2 x \right)}}{\cos{\left(2 x \right)}}\, dx
Integral(sin(2*x)/cos(2*x), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=2xu = 2 x.

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

      sin(u)2cos(u)du\int \frac{\sin{\left(u \right)}}{2 \cos{\left(u \right)}}\, du

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

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

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

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

          (1u)du\int \left(- \frac{1}{u}\right)\, du

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

            1udu=1udu\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du

            1. The integral of 1u\frac{1}{u} is log(u)\log{\left(u \right)}.

            So, the result is: log(u)- \log{\left(u \right)}

          Now substitute uu back in:

          log(cos(u))- \log{\left(\cos{\left(u \right)} \right)}

        So, the result is: log(cos(u))2- \frac{\log{\left(\cos{\left(u \right)} \right)}}{2}

      Now substitute uu back in:

      log(cos(2x))2- \frac{\log{\left(\cos{\left(2 x \right)} \right)}}{2}

    Method #2

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

      2sin(x)cos(x)cos(2x)dx=2sin(x)cos(x)cos(2x)dx\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\cos{\left(2 x \right)}}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\cos{\left(2 x \right)}}\, dx

      1. Rewrite the integrand:

        sin(x)cos(x)cos(2x)=sin(x)cos(x)2cos2(x)1\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\cos{\left(2 x \right)}} = \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{2 \cos^{2}{\left(x \right)} - 1}

      2. Let u=2cos2(x)1u = 2 \cos^{2}{\left(x \right)} - 1.

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

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

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

          1udu=1udu4\int \frac{1}{u}\, du = - \frac{\int \frac{1}{u}\, du}{4}

          1. The integral of 1u\frac{1}{u} is log(u)\log{\left(u \right)}.

          So, the result is: log(u)4- \frac{\log{\left(u \right)}}{4}

        Now substitute uu back in:

        log(2cos2(x)1)4- \frac{\log{\left(2 \cos^{2}{\left(x \right)} - 1 \right)}}{4}

      So, the result is: log(2cos2(x)1)2- \frac{\log{\left(2 \cos^{2}{\left(x \right)} - 1 \right)}}{2}

  2. Add the constant of integration:

    log(cos(2x))2+constant- \frac{\log{\left(\cos{\left(2 x \right)} \right)}}{2}+ \mathrm{constant}


The answer is:

log(cos(2x))2+constant- \frac{\log{\left(\cos{\left(2 x \right)} \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                               
 |                                
 | sin(2*x)          log(cos(2*x))
 | -------- dx = C - -------------
 | cos(2*x)                2      
 |                                
/                                 
sin(2x)cos(2x)dx=Clog(cos(2x))2\int \frac{\sin{\left(2 x \right)}}{\cos{\left(2 x \right)}}\, dx = C - \frac{\log{\left(\cos{\left(2 x \right)} \right)}}{2}
The graph
0.001.000.100.200.300.400.500.600.700.800.90-200000100000
The answer [src]
nan
NaN\text{NaN}
=
=
nan
NaN\text{NaN}
nan
Numerical answer [src]
-0.96117546625349
-0.96117546625349
The graph
Integral of sin(2*x)/cos(2*x) dx

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