All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 143051 by mnjuly1970 last updated on 09/Jun/21
∗∗∗∗∗::LobachevskyIntegral::∗∗∗∗∗ϕ:=∫0∞sin2(tan(x))x2dx=?π2..........
Answered by Olaf_Thorendsen last updated on 09/Jun/21
Letf(x)=sin2(tanx)sin2xϕ=∫0∞sin2(tanx)x2dxϕ=∫0∞sin2(x)x2f(x)dx∀x⩾0fisacontinuousfunctionsatisfyingtheπ−periodicassumptionf(x+π)=f(x),andf(x−π)=f(x).WecanapplytheLobatchevsky′sDirichletintegralformula:∫0∞sin2xx2f(x)dx=∫0∞sinxxf(x)dx=∫0π2f(x)dxϕ=∫0π2sin2(tanx)sin2xdxLetu=tanxϕ=∫0∞sin2usin2(arctanu).du1+u2ϕ=∫0∞sin2u1−cos2(arctanu).du1+u2ϕ=∫0∞sin2u1−11+tan2(arctanu).du1+u2ϕ=∫0∞sin2u1−11+u2.du1+u2ϕ=∫0∞sin2uu2duLetg(u)=1(constantfunctionunity)ϕ=∫0∞sin2uu2g(u)du=∫0∞sinuug(u)du=∫0π2g(u)duϕ=∫0π21.du=π2
Commented by mnjuly1970 last updated on 09/Jun/21
thanksalotmrolaf....
Terms of Service
Privacy Policy
Contact: info@tinkutara.com