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[Post New] 8 Jul 2009 21:32:45 IST
   Subject: Need help :)
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akki ~~ unlucky forever ~~ (1635)

Olaaa!! Perrrfect answer. 275  [405 rates]
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put\ x=sin^2\theta \\ \\ \\ \\ I=2\int_{0}^{\frac{\pi}{2}}(\frac{1}{(1+sin^2\theta)}-\frac{1}{(2+sin^2\theta)})d\theta \\ \\ \\ \\ I = \frac{\pi(\sqrt{3}-1)}{\sqrt6}
all's well that end's well, but if it dosen't, then it is not the end . . .
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