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Question Number 31070 by abdo imad last updated on 02/Mar/18

calculate ∫_0 ^π (dx/(1+2cosx)) .

calculate∫0πdx1+2cosx.

Commented by abdo imad last updated on 03/Mar/18

the ch.tan((x/2))=t give I= ∫_0 ^(+∞) (1/(1+2((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^(+∞) ((2dt)/(1+t^2 +2(1−t^2 ))) = ∫_0 ^(+∞) ((2dt)/(3−t^2 )) =(1/(2(√3)))∫_0 ^(+∞) 2( (1/((√3) −t)) +(1/((√3) +t)))dt =(1/(√3))[ln((√3) +t)−ln((√3) −t)]_0 ^(+∞) =(1/(√3))[ln(∣(((√3) +t)/((√3) −t)))∣]_0 ^(+∞) ⇒ I=0

thech.tan(x2)=tgiveI=∫0+∞11+21−t21+t22dt1+t2=∫0+∞2dt1+t2+2(1−t2)=∫0+∞2dt3−t2=123∫0+∞2(13−t+13+t)dt=13[ln(3+t)−ln(3−t)]0+∞=13[ln(∣3+t3−t)∣]0+∞⇒I=0

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