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StupidGenius
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1355422740273:dw

darthjavier
 2 years ago
Best ResponseYou've already chosen the best response.0Do you have to approximate the answer? Because its integrate indefinite is not an easy function...

across
 2 years ago
Best ResponseYou've already chosen the best response.0First 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.

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

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

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

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