Let I = \(\int\limits_0^{\pi} \frac{x\, tan x}{sec x \,cosec x}dx\) = \(\int\limits_0^{\pi} \cfrac{x\, \frac{sinx}{cosx}}{\cfrac{1}{sinx cosx}}dx\)
(\(\because\) tan = \(\frac{sin x}{cos x}, sec x = \frac{1}{cos x} and\, cosec\, x = \frac{1}{sinx}\) )
Hence, = \(\int\limits_0^{\pi} \frac{x\, tan x}{sec x \,cosec x}dx\) = \(\frac{\pi^2}{4}\)