Question

What is the definite integral of log (tan x) with range from 90 to 0

Answer 1

Substituting x with π/2 - x gives the equivalent integral,

$\displaystyle\int_0^{\frac\pi2}\log(\tan(x))\,\mathrm dx=-\int_{\frac\pi2}^0\log(\cot(x))(-\mathrm dx)=\int_0^{\frac\pi2}\log(\cot(x))\,\mathrm dx$

So if we let J denote the value of the integral, we have

$J=\displaystyle\int_0^{\frac\pi2}\log(\tan (x))\,\mathrm dx$

$J=\displaystyle\int_0^{\frac\pi2}\log(\cot (x))\,\mathrm dx$

$\implies 2J=\displaystyle\int_0^{\frac\pi2}\left(\log(\tan (x))+\log(\cot (x))\right)\,\mathrm dx$

Condensing the logarithms, we have

log(tan(x)) + log(cot(x)) = log(tan(x) cot(x)) = log(1) = 0

since cot(x) = 1/tan(x), which means

$2J=\displaystyle\int_0^{\frac\pi2}0\,\mathrm dx=0$

and so the original integral has a value of J = 0.