solve

and

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- June 21st 2009, 11:55 PMsimplependulumChallenging Integral
solve

and

- June 22nd 2009, 06:52 AMNonCommAlg
i'll solve the first one only and i'll leave the second integral to you or whoever likes trouble! first we need the following magical identity: (Wink)

now finding is straightforward. to find substitute: then and thus

finally let and to get: note that now everything is nice and easy and you're good to go!

**Remark**: i think this is true that for any we have where is some rational function of .

actually i think where both and are monic polynomials with integer coefficients! for example, in your problem: i also think that we

always have these claims are either wrong or easy to prove! (Rofl) ... let's just leave it as a challenge for whoever has time and patience to try it.

by the way, the identity shows that my claims are not based on nothing! (Evilgrin) - June 22nd 2009, 09:29 PMsimplependulum
My solution is to substitute

so,

Although it looks very troublesome , it become quite well after separating it into fractions .

However , i think your method is better than what i am using . (Clapping)