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0-2pi-dx-1-sinxcosx-4piln2-3-




Question Number 216695 by sniper237 last updated on 16/Feb/25
∫_0 ^(2π) (dx/(1+sinxcosx))=^?  ((4πln2)/( (√3)))
02πdx1+sinxcosx=?4πln23
Answered by Ghisom last updated on 16/Feb/25
∫_0 ^(2π) (dx/(1+sin x cos x))=8∫_(π/4) ^(3π/4) (dx/(2+sin 2x))=  =8∫_0 ^(π/2) (dx/(2+cos 2x))=       [t=tan x]  =8∫_0 ^∞ (dt/(t^2 +3))=((8(√3))/3)[arctan (((√3)t)/3)]_0 ^∞ =  =((4π(√3))/3)
2π0dx1+sinxcosx=83π/4π/4dx2+sin2x==8π/20dx2+cos2x=[t=tanx]=80dtt2+3=833[arctan3t3]0==4π33

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