Question Number 102366 by bobhans last updated on 08/Jul/20

∫_0 ^(π/2)  ((cos x)/(1+cos x+sin x)) dx ?

Answered by PRITHWISH SEN 2 last updated on 08/Jul/20

I=∫_0 ^(π/2) ((sin x)/(1+cos x+sin x))dx  2I=∫_0 ^(π/2) {1−(1/(1+cos x+sin x))}dx    = (π/2)−(1/2)∫_0 ^(π/2) ((sec^2 (x/2))/(1+tan (x/2)))dx  = (π/2)−ln∣1+tan (x/2)∣_0 ^(π/2)   = (π/2)−ln∣2∣  ∴ I= (𝛑/4) − (1/2)ln2     please check

Commented byDwaipayan Shikari last updated on 08/Jul/20

It is right

Commented byPRITHWISH SEN 2 last updated on 08/Jul/20

sorry i made a mistake . thank you.

Commented bybobhans last updated on 09/Jul/20

yes thank you

Answered by Dwaipayan Shikari last updated on 08/Jul/20

∫_0 ^(π/2) ((cosx)/(1+cosx+sinx))=∫((sinx)/(1+cosx+sinx))=I  2I=∫_0 ^(π/2) 1−(1/(1+cosx+sinx))dx  2I=(π/2)−2∫_0 ^1 (1/(1+((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 )))).(1/(1+t^2 ))dt    put   tan(x/2)=t  2I=(π/2)−2∫_0 ^1 (1/(2(1+t)))dt  2I=(π/2)−[log(1+t)]_0 ^1   2I=(π/2)−log2  I=(π/4)−(1/2)log2

Answered by mathmax by abdo last updated on 08/Jul/20

I =∫_0 ^(π/2)  ((cosx)/(1+cosx +sinx))dx cha7gement tan((x/2))=t give  I =∫_0 ^1  (((1−t^2 )/(1+t^2 ))/(1+((1−t^2 )/(1+t^2 )) +((2t)/(1+t^2 ))))×((2dt)/(1+t^2 )) =2∫_0 ^1  ((1−t^2 )/((1+t^2  +1−t^2  +2t)(t^2 +1)))dt  =∫_0 ^1  ((1−t^2 )/((t+1)(t^2 +1)))dt =∫_0 ^(1 ) ((1−t)/(t^2  +1))dt =∫_0 ^1  (dt/(t^2  +1)) −(1/2)∫_0 ^1  ((2t)/(t^2  +1))dt  =[arctan(t)]_0 ^1  −(1/2) [ln(t^2  +1)]_0 ^1  =(π/4)−((ln(2))/2)