1. ## Another interesting integral

evaluate the integral $\displaystyle \int_{0}^{\frac{\pi}{2}} x \cot(x) ~dx$

In fact , it is not a challenge problem for an experienced integrator
but i believe that this may help remind us of some techniques

2. Originally Posted by simplependulum
evaluate the integral $\displaystyle \int_{0}^{\frac{\pi}{2}} x \cot(x) ~dx$

In fact , it is not a challenge problem for an experienced integrator
but i believe that this may help remind us of some techniques
let $\displaystyle x=u$ and $\displaystyle \cot x \ dx = dv.$ then integration by parts gives us: $\displaystyle I=\int_0^{\frac{\pi}{2}} x \cot x \ dx = -\int_0^{\frac{\pi}{2}} \ln(\sin x) \ dx.$ thus:
$\displaystyle -I=\frac{\pi}{2} \ln 2 + \int_0^{\frac{\pi}{2}} \ln(\sin(x/2)) \ dx + \int_0^{\frac{\pi}{2}} \ln(\cos(x/2)) \ dx$

$\displaystyle =\frac{\pi}{2} \ln 2 + \int_0^{\frac{\pi}{2}} \ln(\sin(x/2)) \ dx + \int_{\frac{\pi}{2}}^{\pi} \ln(\sin(x/2)) \ dx$

$\displaystyle =\frac{\pi}{2} \ln 2 + \int_0^{\pi} \ln(\sin(x/2)) \ dx=\frac{\pi}{2} \ln 2 + 2\int_0^{\frac{\pi}{2}} \ln(\sin x) \ dx$

$\displaystyle =\frac{\pi}{2} \ln 2 - 2I.$ therefore: $\displaystyle I=\frac{\pi}{2}\ln 2.$

3. terrific !