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Question Number 36439 by prof Abdo imad last updated on 02/Jun/18

calculate ∫_0 ^π ((sin(2x))/(2 +cosx))dx

calculate∫0πsin(2x)2+cosxdx

Commented by abdo.msup.com last updated on 03/Jun/18

I = ∫_0 ^π ((2sinx cosx)/(2+cosx)) dx changement cosx =t give −sinx dx =dt I = ∫_1 ^(−1) ((−2t dt)/(2+t)) =2 ∫_(−1) ^1 (t/(t+2))dt =2 ∫_(−1) ^1 ((t+2 −2)/(t+2))dt =4 −4 [ln∣t+2∣]_(−1) ^1 =4 −4( ln(3)) I =4 −4ln(3) .

I=∫0π2sinxcosx2+cosxdxchangementcosx=tgive−sinxdx=dtI=∫1−1−2tdt2+t=2∫−11tt+2dt=2∫−11t+2−2t+2dt=4−4[ln∣t+2∣]−11=4−4(ln(3))I=4−4ln(3).

Answered by ajfour last updated on 02/Jun/18

let cos x=t I =∫_1 ^( −1) ((−2tdt)/(2+t)) =−2∫_1 ^( −1) dt+4∫_1 ^( −1) (dt/(t+2)) I = 4−ln 3 .

letcosx=tI=∫1−1−2tdt2+t=−2∫1−1dt+4∫1−1dtt+2I=4−ln3.

Commented by MJS last updated on 02/Jun/18

minor mistake at the end: I=4−4ln 3

minormistakeattheend:I=4−4ln3

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