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Question Number 110551 by shahria14 last updated on 29/Aug/20

Answered by Dwaipayan Shikari last updated on 29/Aug/20

∫_0 ^π (x/(1+sinx))dx=∫_0 ^π ((π−x)/(1+sinx))dx=I  2I=∫_0 ^π (π/(1+sinx))dx=2π∫_0 ^∞ (1/(1+((2t)/(1+t^2 )))).(1/(1+t^2 ))dt        (tan(x/2)=t)  2I=2π∫_0 ^∞ (1/((1+t)^2 ))dt  I=−π[(1/(1+t))]_0 ^∞ =π

0πx1+sinxdx=0ππx1+sinxdx=I2I=0ππ1+sinxdx=2π011+2t1+t2.11+t2dt(tanx2=t)2I=2π01(1+t)2dtI=π[11+t]0=π

Answered by mathmax by abdo last updated on 30/Aug/20

I =∫_0 ^π  ((xdx)/(1+sinx)) changement x =π−t give   I =∫_0 ^π   ((π−t)/(1+sint)) dt =π ∫_0 ^(π )  (dt/(1+sint)) −I ⇒2I =π∫_0 ^π  (dt/(1+sint))  =_(tan((t/2))=u)     ∫_0 ^∞     ((2du)/((1+u^2 )(1+((2u)/(1+u^2 ))))) =∫_0 ^∞   ((2du)/(1+u^2  +2u)) =2∫_0 ^∞   (du/((1+u)^2 ))  =[((−2)/(1+u))]_0 ^∞  =1 ⇒2I =π ⇒I =(π/2)

I=0πxdx1+sinxchangementx=πtgiveI=0ππt1+sintdt=π0πdt1+sintI2I=π0πdt1+sint=tan(t2)=u02du(1+u2)(1+2u1+u2)=02du1+u2+2u=20du(1+u)2=[21+u]0=12I=πI=π2

Commented by mathmax by abdo last updated on 30/Aug/20

sorry 2I =2π ⇒ I =π

sorry2I=2πI=π

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