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Question Number 33494 by mondodotto@gmail.com last updated on 17/Apr/18
∫exsin2xdx
Commented by math khazana by abdo last updated on 18/Apr/18
∫exsin2xdx=∫ex1−cos(2x)2dx=2∫ex1−cos(2x)dx=2x=t2∫et21−costdt2
∫exsin2xdx=∫et2(∑n=0∞cosnt)dt=∑n=0∞∫et2cosntdt=∑n=0∞AnwithAn=∫et2cosntdtintegralcalculablebyrecurence
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