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Question Number 102701 by 175mohamed last updated on 10/Jul/20
Evaluate:∫sinxxdx
Commented by M±th+et+s last updated on 10/Jul/20
gotoQ.98713isolvedthisquestion
Answered by PRITHWISH SEN 2 last updated on 10/Jul/20
∫1x{x−x33!+x55!−...}dx=∑∞r=1(−1)r+1x2r−1(2r−1).(2r−1)!pleasecheck.
Answered by mathmax by abdo last updated on 10/Jul/20
atformofseriewehavesinx=∑n=0∞(−1)nx2n+1(2n+1)!withradiusr=+∞⇒sinxx=∑n=0∞(−1)nx2n(2n+1)!⇒∫sinxxdx=∑n=0∞(−1)n(2n+1)!∫x2ndx=∑n=0∞(−1)n(2n+1)!(2n+1)x2n+1+C
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