RE: Can you solve this integral from MIT's Integration Bee? by mike00632
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mathematics·@rycharde·
0.000 HBDThe key step is using sin(2x) = 2.sinx.cosx so that cos(x) = sin(2x)/2sin(x) and by extension cos(nx) = sin(2nx)/2sin(nx) when plugged into the integral product, all the sines cancel out apart from the first and last terms.