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

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

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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:

            ∫1u du=−∫1u du\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=2∫sin⁡(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)2cos⁡2(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=2cos⁡2(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:

          ∫1u du=−∫1u du4\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⁡(2cos⁡2(x)−1)4- \frac{\log{\left(2 \cos^{2}{\left(x \right)} - 1 \right)}}{4}

      So, the result is: −log⁡(2cos⁡2(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=C−log⁡(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}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-200000100000
Plot the graph f(x)
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.

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