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(1+tgx)/sin2x
Solving integrals/ (1+tgx)/sin2x

Integral of (1+tgx)/sin2x dx

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  0              
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 |  1 + tan(x)   
 |  ---------- dx
 |   sin(2*x)    
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∫00tan⁡(x)+1sin⁡(2x) dx\int\limits_{0}^{0} \frac{\tan{\left(x \right)} + 1}{\sin{\left(2 x \right)}}\, dx 
Integral((1 + tan(x))/sin(2*x), (x, 0, 0))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        tan⁡(x)sin⁡(2x)=tan⁡(x)2sin⁡(x)cos⁡(x)\frac{\tan{\left(x \right)}}{\sin{\left(2 x \right)}} = \frac{\tan{\left(x \right)}}{2 \sin{\left(x \right)} \cos{\left(x \right)}}

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

        ∫tan⁡(x)2sin⁡(x)cos⁡(x) dx=∫tan⁡(x)sin⁡(x)cos⁡(x) dx2\int \frac{\tan{\left(x \right)}}{2 \sin{\left(x \right)} \cos{\left(x \right)}}\, dx = \frac{\int \frac{\tan{\left(x \right)}}{\sin{\left(x \right)} \cos{\left(x \right)}}\, dx}{2}

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

          But the integral is

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

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

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

        But the integral is

        log⁡(cos⁡(2x)−1)4−log⁡(cos⁡(2x)+1)4\frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{4} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{4}

      The result is: log⁡(cos⁡(2x)−1)4−log⁡(cos⁡(2x)+1)4+sin⁡(x)2cos⁡(x)\frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{4} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{4} + \frac{\sin{\left(x \right)}}{2 \cos{\left(x \right)}}

    Method #2

    1. Rewrite the integrand:

      tan⁡(x)+1sin⁡(2x)=tan⁡(x)2sin⁡(x)cos⁡(x)+12sin⁡(x)cos⁡(x)\frac{\tan{\left(x \right)} + 1}{\sin{\left(2 x \right)}} = \frac{\tan{\left(x \right)}}{2 \sin{\left(x \right)} \cos{\left(x \right)}} + \frac{1}{2 \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:

        ∫tan⁡(x)2sin⁡(x)cos⁡(x) dx=∫tan⁡(x)sin⁡(x)cos⁡(x) dx2\int \frac{\tan{\left(x \right)}}{2 \sin{\left(x \right)} \cos{\left(x \right)}}\, dx = \frac{\int \frac{\tan{\left(x \right)}}{\sin{\left(x \right)} \cos{\left(x \right)}}\, dx}{2}

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

          But the integral is

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

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

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

        ∫12sin⁡(x)cos⁡(x) dx=∫1sin⁡(x)cos⁡(x) dx2\int \frac{1}{2 \sin{\left(x \right)} \cos{\left(x \right)}}\, dx = \frac{\int \frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}}\, dx}{2}

        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)}

        So, the result is: −log⁡(sin⁡2(x)−1)4+log⁡(sin⁡(x))2- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{4} + \frac{\log{\left(\sin{\left(x \right)} \right)}}{2}

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

  2. Now simplify:

    log⁡(−sin⁡2(x))4−log⁡(cos⁡2(x))4+tan⁡(x)2\frac{\log{\left(- \sin^{2}{\left(x \right)} \right)}}{4} - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{4} + \frac{\tan{\left(x \right)}}{2}

  3. Add the constant of integration:

    log⁡(−sin⁡2(x))4−log⁡(cos⁡2(x))4+tan⁡(x)2+constant\frac{\log{\left(- \sin^{2}{\left(x \right)} \right)}}{4} - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{4} + \frac{\tan{\left(x \right)}}{2}+ \mathrm{constant}

The answer is:

log⁡(−sin⁡2(x))4−log⁡(cos⁡2(x))4+tan⁡(x)2+constant\frac{\log{\left(- \sin^{2}{\left(x \right)} \right)}}{4} - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{4} + \frac{\tan{\left(x \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                     
 |                                                                      
 | 1 + tan(x)          log(1 + cos(2*x))   log(-1 + cos(2*x))    sin(x) 
 | ---------- dx = C - ----------------- + ------------------ + --------
 |  sin(2*x)                   4                   4            2*cos(x)
 |                                                                      
/
∫tan⁡(x)+1sin⁡(2x) dx=C+log⁡(cos⁡(2x)−1)4−log⁡(cos⁡(2x)+1)4+sin⁡(x)2cos⁡(x)\int \frac{\tan{\left(x \right)} + 1}{\sin{\left(2 x \right)}}\, dx = C + \frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{4} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{4} + \frac{\sin{\left(x \right)}}{2 \cos{\left(x \right)}}
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Integral of (1+tgx)/sin2x dx

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

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