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tan((pi)(x)/2)
Solving integrals/ tan((pi)(x)/2)

Integral of tan((pi)(x)/2) dx

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  1             
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∫01tan⁡(πx2) dx\int\limits_{0}^{1} \tan{\left(\frac{\pi x}{2} \right)}\, dx 
Integral(tan(pi*x/2), (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    tan⁡(πx2)=sin⁡(πx2)cos⁡(πx2)\tan{\left(\frac{\pi x}{2} \right)} = \frac{\sin{\left(\frac{\pi x}{2} \right)}}{\cos{\left(\frac{\pi x}{2} \right)}}

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=cos⁡(πx2)u = \cos{\left(\frac{\pi x}{2} \right)}.

      Then let du=−πsin⁡(πx2)dx2du = - \frac{\pi \sin{\left(\frac{\pi x}{2} \right)} dx}{2} and substitute −2duπ- \frac{2 du}{\pi}:

      ∫4π2u du\int \frac{4}{\pi^{2} u}\, du

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

        ∫(−2uπ) du=−2∫1u duπ\int \left(- \frac{2}{u \pi}\right)\, du = - \frac{2 \int \frac{1}{u}\, du}{\pi}

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

        So, the result is: −2log⁡(u)π- \frac{2 \log{\left(u \right)}}{\pi}

      Now substitute uu back in:

      −2log⁡(cos⁡(πx2))π- \frac{2 \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}}{\pi}

    Method #2

    1. Let u=πx2u = \frac{\pi x}{2}.

      Then let du=πdx2du = \frac{\pi dx}{2} and substitute 2duπ\frac{2 du}{\pi}:

      ∫4sin⁡(u)π2cos⁡(u) du\int \frac{4 \sin{\left(u \right)}}{\pi^{2} \cos{\left(u \right)}}\, du

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

        ∫2sin⁡(u)πcos⁡(u) du=2∫sin⁡(u)cos⁡(u) duπ\int \frac{2 \sin{\left(u \right)}}{\pi \cos{\left(u \right)}}\, du = \frac{2 \int \frac{\sin{\left(u \right)}}{\cos{\left(u \right)}}\, du}{\pi}

        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 \frac{1}{u}\, du

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

            ∫(−1u) du=−∫1u du\int \left(- \frac{1}{u}\right)\, 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: −2log⁡(cos⁡(u))π- \frac{2 \log{\left(\cos{\left(u \right)} \right)}}{\pi}

      Now substitute uu back in:

      −2log⁡(cos⁡(πx2))π- \frac{2 \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}}{\pi}

  3. Add the constant of integration:

    −2log⁡(cos⁡(πx2))π+constant- \frac{2 \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}}{\pi}+ \mathrm{constant}

The answer is:

−2log⁡(cos⁡(πx2))π+constant- \frac{2 \log{\left(\cos{\left(\frac{\pi x}{2} \right)} \right)}}{\pi}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                        /   /pi*x\\
 |                    2*log|cos|----||
 |    /pi*x\               \   \ 2  //
 | tan|----| dx = C - ----------------
 |    \ 2  /                 pi       
 |                                    
/
2 log⁡sec⁡(π x2)π{{2\,\log \sec \left({{\pi\,x}\over{2}}\right)}\over{\pi}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90020000000000000000
Plot the graph f(x)
The answer [src] 
oo
−2 log⁡∣cos⁡(π2)∣π-{{2\,\log \left| \cos \left({{\pi}\over{2}}\right)\right| }\over{ \pi}} 
=
= 
oo
∞\infty
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Numerical answer [src] 
27.7822258859063
27.7822258859063
The graph
Integral of tan((pi)(x)/2) dx

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

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