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Question Number 91018 by M±th+et+s last updated on 27/Apr/20
∫0πsin21x2sinx2dx
Commented by mathmax by abdo last updated on 29/Apr/20
lettakeatrychangementx2=tgive∫0πsin(21x2)sin(x2)dx=2∫0π2sin(21t)sintdtletw(t)=1sintwehavew(t)=2eit−e−it=2z−z−1(z=eit)=2zz2−1=−2z(11−z2)=−2z∑n=0∞z2n=−2eit∑n=0∞e2int⇒I=−4∫0π2sin(21t)∑n=0∞e(2n+1)itdt=−4∑n=0∞∫0π2sin(21t)e(2n+1)tdt=−4∑n=0∞AnAn=∫0π2sin(21t)(cos(2n+1)t+isin(2n+1)t)dt=∫0π2sin(21t)cos(2n+1)t+i∫0π2sin(21t)sin(2n+1)tdtsinacosb=cos(π2−a)cosb=12(cos(π2−a+b)+cos(π2−a−b))=12{sin(a−b)+sin(a+b)}⇒∫0π2sin(21t)cos(2n+1)tdt=12∫0π2{sin(20−2n)t+sin(2n+22)t}dt=12[−120−2ncos(20−2n)t+12n+22cos(2n+22)t]0π2=12{−120−2ncos(10−n)π+12n+22cos(n+11)π+120−2n−12n+22}....becontinued...
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