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Question Number 97173 by 675480065 last updated on 06/Jun/20
Answered by Sourav mridha last updated on 06/Jun/20
I=2π∫−π4+π4dx(1+esinx)(2−cos2x)...(i)andalsoI=2π∫+π4−π4d(−x)(1+esin(−x))(2−cos2(−x))=2π∫−π4+π4esin(x)(1+esin(x))(2−cos2x)dx.....(ii)now(i)+(ii)weget,2I=2π∫−π4+π4dx2sin2(x)+1[evenfn]soI=−2π∫0+π4d(cotx)(3)2+(cot(x))2=−2π.13[tan−1(cotx3)]0+π4=−2π.13[π6−π2]=233sonowI2=427⇒27I2=4
Commented by 675480065 last updated on 06/Jun/20
idontunderstandfrom(i)+(ik)
Commented by Sourav mridha last updated on 07/Jun/20
trytodostepbystep−−atfirstadd(i)+(ii)thenyouwillsee(1+esin(x))isvanish.nowitsremain(2−cos2x)atdenomenator.this=12sin2(x)+1=cosec2(x)3+cot2(x)nowgoesonandcheackwithmysoln.
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