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Question Number 214051 by issac last updated on 25/Nov/24
evaluate∫0πesin2(u)du...iuseFeynman′stricktosolveintegral∫0πesin2(u)du=II(t)=∫0πetsin2(u)duI(1)(t)=∫0πsin2(u)etsin2(u)du∫0π(1−cos2(u))et⋅sin2(u)du∫0πet⋅sin2(u)du−∫0πcos2(u)et⋅sin2(u)du∫0πet⋅sin2(u)du−∫0πsin2(u+π2)et⋅sin2(u)duu+π2=w→du=dw∫π23π2sin2(w)et⋅cos2(w)dw=∫π23π2sin2(w)et(1−sin2(w))dw∫π23π2sin2(w)ete−tsin2(w)dw=et∫π23π2sin2(w)e−tsin2(w)dwI(1)(t)=∫0πet⋅sin2(u)du−et∫0πsin2(u)e−t⋅sin2(u)duIcan′t...anymore...help
Commented by Ghisom last updated on 26/Nov/24
Feynmancan′thelphere
Answered by Ghisom last updated on 26/Nov/24
∫π0esin2xdx=2∫π/20esin2xdx=[t=2x+π]=e∫2ππe12costdt=e∫π0e12costdtModifiedBesselFunctionofthe1stKindIn(z)=1π∫π0ezcosθcos(nθ)dθinourcasen=0∧z=12⇒e∫π0e12costdt=eπI0(12)
Commented by issac last updated on 27/Nov/24
wowthx
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