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Solving integrals/ ln(1+tgx)

Integral of ln(1+tgx) dx

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∫01log⁡(tan⁡(x)+1) dx\int\limits_{0}^{1} \log{\left(\tan{\left(x \right)} + 1 \right)}\, dx 
Integral(log(1 + tan(x)), (x, 0, 1))
Detail solution

  1. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(x)=log⁡(tan⁡(x)+1)u{\left(x \right)} = \log{\left(\tan{\left(x \right)} + 1 \right)} and let dv⁡(x)=1\operatorname{dv}{\left(x \right)} = 1.

    Then du⁡(x)=tan⁡2(x)+1tan⁡(x)+1\operatorname{du}{\left(x \right)} = \frac{\tan^{2}{\left(x \right)} + 1}{\tan{\left(x \right)} + 1}.

    To find v(x)v{\left(x \right)}:

    1. The integral of a constant is the constant times the variable of integration:

      ∫1 dx=x\int 1\, dx = x

    Now evaluate the sub-integral.

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

    2. Rewrite the integrand:

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

    3. Integrate term-by-term:

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

        But the integral is

        ∫xtan⁡2(x)tan⁡(x)+1 dx\int \frac{x \tan^{2}{\left(x \right)}}{\tan{\left(x \right)} + 1}\, dx

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

        But the integral is

        ∫xtan⁡(x)+1 dx\int \frac{x}{\tan{\left(x \right)} + 1}\, dx

      The result is: ∫xtan⁡(x)+1 dx+∫xtan⁡2(x)tan⁡(x)+1 dx\int \frac{x}{\tan{\left(x \right)} + 1}\, dx + \int \frac{x \tan^{2}{\left(x \right)}}{\tan{\left(x \right)} + 1}\, dx

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

        But the integral is

        ∫xtan⁡2(x)tan⁡(x)+1 dx\int \frac{x \tan^{2}{\left(x \right)}}{\tan{\left(x \right)} + 1}\, dx

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

        But the integral is

        ∫xtan⁡(x)+1 dx\int \frac{x}{\tan{\left(x \right)} + 1}\, dx

      The result is: ∫xtan⁡(x)+1 dx+∫xtan⁡2(x)tan⁡(x)+1 dx\int \frac{x}{\tan{\left(x \right)} + 1}\, dx + \int \frac{x \tan^{2}{\left(x \right)}}{\tan{\left(x \right)} + 1}\, dx

  3. Add the constant of integration:

    xlog⁡(tan⁡(x)+1)−∫xtan⁡(x)+1 dx−∫xtan⁡2(x)tan⁡(x)+1 dx+constantx \log{\left(\tan{\left(x \right)} + 1 \right)} - \int \frac{x}{\tan{\left(x \right)} + 1}\, dx - \int \frac{x \tan^{2}{\left(x \right)}}{\tan{\left(x \right)} + 1}\, dx+ \mathrm{constant}

The answer is:

xlog⁡(tan⁡(x)+1)−∫xtan⁡(x)+1 dx−∫xtan⁡2(x)tan⁡(x)+1 dx+constantx \log{\left(\tan{\left(x \right)} + 1 \right)} - \int \frac{x}{\tan{\left(x \right)} + 1}\, dx - \int \frac{x \tan^{2}{\left(x \right)}}{\tan{\left(x \right)} + 1}\, dx+ \mathrm{constant}

The answer (Indefinite) [src] 
                                                 /                                 
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  /                          |                  |      2                           
 |                           |     x            | x*tan (x)                        
 | log(1 + tan(x)) dx = C -  | ---------- dx -  | ---------- dx + x*log(1 + tan(x))
 |                           | 1 + tan(x)       | 1 + tan(x)                       
/                            |                  |                                  
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x log⁡(tan⁡x+1)−x log⁡(tan⁡2x+2 tan⁡x+12)−atan2(tan⁡x+12,tan⁡x+12) log⁡(tan⁡2x+1)−i li2(i ((i+1) tan⁡x−i+1)2)+i li2(i ((i−1) tan⁡x−i−1)2)2x\,\log \left(\tan x+1\right)-{{x\,\log \left({{\tan ^2x+2\,\tan x+ 1}\over{2}}\right)-{\rm atan2}\left({{\tan x+1}\over{2}} , {{\tan x+ 1}\over{2}}\right)\,\log \left(\tan ^2x+1\right)-i\,{\it li}_{2}({{i \,\left(\left(i+1\right)\,\tan x-i+1\right)}\over{2}})+i\,{\it li}_{ 2}({{i\,\left(\left(i-1\right)\,\tan x-i-1\right)}\over{2}})}\over{2 }}
Simplify
The answer [src] 
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−4 log⁡(tan⁡21+2 tan⁡1+12)−π log⁡(tan⁡21+1)+4 i li2(−(i+1) tan⁡1+i−12)−4 i li2((i−1) tan⁡1+i+12)8+log⁡(tan⁡1+1)−i li2(i+12)−i li2(−i−12)2-{{4\,\log \left({{\tan ^21+2\,\tan 1+1}\over{2}}\right)-\pi\,\log \left(\tan ^21+1\right)+4\,i\,{\it li}_{2}(-{{\left(i+1\right)\, \tan 1+i-1}\over{2}})-4\,i\,{\it li}_{2}({{\left(i-1\right)\,\tan 1+ i+1}\over{2}})}\over{8}}+\log \left(\tan 1+1\right)-{{i\,{\it li}_{2 }({{i+1}\over{2}})-i\,{\it li}_{2}(-{{i-1}\over{2}})}\over{2}} 
=
= 
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∫01log⁡(tan⁡(x)+1) dx\int\limits_{0}^{1} \log{\left(\tan{\left(x \right)} + 1 \right)}\, dx
Simplify
Numerical answer [src] 
0.446043170167152
0.446043170167152

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

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