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Solving integrals/ (1+sin4x)/(sin2x)^2

Integral of (1+sin4x)/(sin2x)^2 dx

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  1                
  /                
 |                 
 |  1 + sin(4*x)   
 |  ------------ dx
 |      2          
 |   sin (2*x)     
 |                 
/                  
0
∫01sin⁡(4x)+1sin⁡2(2x) dx\int\limits_{0}^{1} \frac{\sin{\left(4 x \right)} + 1}{\sin^{2}{\left(2 x \right)}}\, dx 
Integral((1 + sin(4*x))/sin(2*x)^2, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      1. Rewrite the integrand:

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

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

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

            Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx 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⁡(x))- \log{\left(\cos{\left(x \right)} \right)}

          So, the result is: 2log⁡(cos⁡(x))2 \log{\left(\cos{\left(x \right)} \right)}

        1. Don't know the steps in finding this integral.

          But the integral is

          −log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)}

        The result is: −log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))+2log⁡(cos⁡(x))- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)}

      1. Don't know the steps in finding this integral.

        But the integral is

        −cos⁡(2x)2sin⁡(2x)- \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}

      The result is: −log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))+2log⁡(cos⁡(x))−cos⁡(2x)2sin⁡(2x)- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}

    Method #2

    1. Rewrite the integrand:

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

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

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

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx 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⁡(x))- \log{\left(\cos{\left(x \right)} \right)}

        So, the result is: 2log⁡(cos⁡(x))2 \log{\left(\cos{\left(x \right)} \right)}

      1. Don't know the steps in finding this integral.

        But the integral is

        −log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)}

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

        ∫14sin⁡2(x)cos⁡2(x) dx=∫1sin⁡2(x)cos⁡2(x) dx4\int \frac{1}{4 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx = \frac{\int \frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx}{4}

        1. Don't know the steps in finding this integral.

          But the integral is

          −2cos⁡(2x)sin⁡(2x)- \frac{2 \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}

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

      The result is: −log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))+2log⁡(cos⁡(x))−cos⁡(2x)2sin⁡(2x)- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}

  2. Now simplify:

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

  3. Add the constant of integration:

    −log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))+2log⁡(cos⁡(x))−12tan⁡(2x)+constant- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{1}{2 \tan{\left(2 x \right)}}+ \mathrm{constant}

The answer is:

−log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))+2log⁡(cos⁡(x))−12tan⁡(2x)+constant- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{1}{2 \tan{\left(2 x \right)}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                  
 |                                          /        2   \                           
 | 1 + sin(4*x)                          log\-1 + sin (x)/    cos(2*x)               
 | ------------ dx = C + 2*log(cos(x)) - ----------------- - ---------- + log(sin(x))
 |     2                                         2           2*sin(2*x)              
 |  sin (2*x)                                                                        
 |                                                                                   
/
∫sin⁡(4x)+1sin⁡2(2x) dx=C−log⁡(sin⁡2(x)−1)2+log⁡(sin⁡(x))+2log⁡(cos⁡(x))−cos⁡(2x)2sin⁡(2x)\int \frac{\sin{\left(4 x \right)} + 1}{\sin^{2}{\left(2 x \right)}}\, dx = C - \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}
Simplify
The answer [src] 
     pi*I
oo - ----
      2
∞−iπ2\infty - \frac{i \pi}{2} 
=
= 
     pi*I
oo - ----
      2
∞−iπ2\infty - \frac{i \pi}{2} 
oo - pi*i/2
Numerical answer [src] 
3.44830919487149e+18
3.44830919487149e+18

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

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