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StupidGeniusBest ResponseYou've already chosen the best response.0
dw:1355422740273:dw
 one year ago

darthjavierBest ResponseYou've already chosen the best response.0
Do you have to approximate the answer? Because its integrate indefinite is not an easy function...
 one year ago

acrossBest ResponseYou've already chosen the best response.0
First find the antiderivative of \(\sin^2t\):\[\int\sin^2t\,dt=\frac t2\frac14\sin2t+c,\]then apply the limits:\[\left[\frac t2\frac14\sin2t\right]_0^\infty.\]However, since\[\lim_{t\to\infty}\sin 2t\]doesn't converge, the integral doesn't either.
 one year ago

darthjavierBest ResponseYou've already chosen the best response.0
No, he wrote \[\mathrm{ \int_{0}^\infty \sin(x^2)dx}\]
 one year ago

darthjavierBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=+%5Cint_%7B0%7D%5E%5Cinfty+%5Csin%28x%5E2%29dx
 one year ago

doritoBest ResponseYou've already chosen the best response.0
\[\sum_{0}^{\infty} ((1)^k/(2k+1))*(x^2)^{2k+1}\]
 one year ago

StupidGeniusBest ResponseYou've already chosen the best response.0
thanks for all but i want the steps of solution
 one year ago
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