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find-pi-4-pi-4-cosx-2-5sinx-dx-




Question Number 65776 by mathmax by abdo last updated on 03/Aug/19
find  ∫_(−(π/4)) ^(π/4)   ((cosx)/(2+5sinx))dx
findπ4π4cosx2+5sinxdx
Commented by kaivan.ahmadi last updated on 04/Aug/19
u=2+5sinx⇒du=5cosxdx  (1/5)∫(du/u)=(1/5)lnu=(1/5)ln(2+5sinx)∣_(−(π/4)) ^(π/4) =
u=2+5sinxdu=5cosxdx15duu=15lnu=15ln(2+5sinx)π4π4=