# Another interesting integral

• Jun 6th 2009, 07:50 PM
simplependulum
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 (Nod)
• Jun 6th 2009, 08:41 PM
NonCommAlg
Quote:

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 (Nod)

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: $\displaystyle I=\frac{\pi}{2}\ln 2.$
• Jun 8th 2009, 02:03 AM
simplependulum
terrific !