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Question Number 214051 by issac last updated on 25/Nov/24

evaluate ∫_0 ^( π)  e^(sin^2 (u)) du...   i use Feynman′s trick to solve integral  ∫_0 ^( π)  e^(sin^2 (u)) du=I  I(t)=∫_0 ^( π)  e^(tsin^2 (u)) du  I^((1)) (t)=∫_0 ^( π)  sin^2 (u)e^(tsin^2 (u)) du  ∫_0 ^( π) (1−cos^2 (u))e^(t∙sin^2 (u)) du  ∫_0 ^( π)  e^(t∙sin^2 (u)) du−∫_0 ^( π)  cos^2 (u)e^(t∙sin^2 (u)) du  ∫_0 ^( π)  e^(t∙sin^2 (u)) du−∫_0 ^( π)  sin^2 (u+(π/2))e^(t∙sin^2 (u)) du  u+(π/2) =w  →  du=dw  ∫_(π/2) ^((3π)/2)  sin^2 (w)e^(t∙cos^2 (w))  dw=  ∫_(π/2) ^( ((3π)/2))  sin^2 (w)e^(t(1−sin^2 (w))) dw  ∫_(π/2) ^((3π)/2)   sin^2 (w)e^t e^(−tsin^2 (w)) dw=e^t ∫_(π/2) ^((3π)/2)  sin^2 (w)e^(−tsin^2 (w)) dw  I^((1)) (t)=∫_0 ^( π)  e^(t∙sin^2 (u)) du−e^t ∫_0 ^( π)  sin^2 (u)e^(−t∙sin^2 (u)) du  I can′t ...anymore...help

evaluate0πesin2(u)du...iuseFeynmanstricktosolveintegral0πesin2(u)du=II(t)=0πetsin2(u)duI(1)(t)=0πsin2(u)etsin2(u)du0π(1cos2(u))etsin2(u)du0πetsin2(u)du0πcos2(u)etsin2(u)du0πetsin2(u)du0πsin2(u+π2)etsin2(u)duu+π2=wdu=dwπ23π2sin2(w)etcos2(w)dw=π23π2sin2(w)et(1sin2(w))dwπ23π2sin2(w)etetsin2(w)dw=etπ23π2sin2(w)etsin2(w)dwI(1)(t)=0πetsin2(u)duet0πsin2(u)etsin2(u)duIcant...anymore...help

Commented by Ghisom last updated on 26/Nov/24

Feynman can′t help here

Feynmancanthelphere

Answered by Ghisom last updated on 26/Nov/24

∫_0 ^π e^(sin^2  x) dx=2∫_0 ^(π/2) e^(sin^2  x) dx=       [t=2x+π]  =(√e)∫_π ^(2π) e^((1/2)cos t) dt=(√e)∫_0 ^π e^((1/2)cos t) dt  Modified Bessel Function of the 1^(st)  Kind  I_n (z)=(1/π)∫_0 ^π e^(zcos θ) cos (nθ) dθ  in our case n=0∧z=(1/2) ⇒  (√e)∫_0 ^π e^((1/2)cos t) dt=(√e)πI_0  ((1/2))

π0esin2xdx=2π/20esin2xdx=[t=2x+π]=e2ππe12costdt=eπ0e12costdtModifiedBesselFunctionofthe1stKindIn(z)=1ππ0ezcosθcos(nθ)dθinourcasen=0z=12eπ0e12costdt=eπI0(12)

Commented by issac last updated on 27/Nov/24

wow thx

wowthx

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