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Question Number 141333 by cesarL last updated on 17/May/21
∫x2cos(x2)dx
Answered by qaz last updated on 17/May/21
∫x2cosx2dx=2x2sinx2−4∫xsinx2dx=2x2sinx2−4(−2xcosx2+2∫cosx2dx)=2x2sinx2+8xcosx2−16sinx2+C
Commented by cesarL last updated on 17/May/21
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