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0-a-1-x-a-2-x-2-dx-




Question Number 6519 by benny last updated on 30/Jun/16
 ∫_( 0) ^a    (1/(x+(√(a^2 −x^2 )))) dx =
a01x+a2x2dx=
Commented by Tawakalitu. last updated on 30/Jun/16
(Π/4)
Π4
Commented by Tawakalitu. last updated on 01/Jul/16
x = a sint   dx = a cost dt   ∴ ∫(dx/(x + (√(a^2  − x^2 ))))   = ∫((a cost dt)/(a sint + (√(a^2  − a^2  sin^2 t))))  = ∫((a cost dt)/(a sint + (√(a^2  − a^2  (1 − cos^2 t))))) = ∫((a cost dt)/(a sint + (√(a^2  − a^2  + a^2 cos^2 t))))  = ∫((a cost dt)/(a sint + (√(a^2  cos^2 t))))  = ∫((a cost dt)/(a sint + a cost))  =  ∫((a cost dt)/(a(sint + cost)))  = ∫((cost dt)/(sint + cost))  = ∫(((1/2) × 2 cost dt)/(sint + cost))  = ∫(((1/2) × (cost + cost) dt)/(sint + cost))  = ∫(((1/2) × (cost + sint + cost − sint) dt)/(sint + cost))   = ∫(((1/2)(cost + sint)dt + (1/2)(cost − sint)dt)/(cost + sint))  = ∫(1/2) × ((cost + sint)/(cost + sint)) dt + ∫(1/2) × ((cost − sint)/(cost + sint)) dt  = ∫(1/2) dt + ∫(1/2) × ((cost − sint)/(cost + sint)) dt  let u = cost + sint,   (du/dt) = − sint + cost , du = (cost − sint) dt  = ∫(1/2) dt + ∫(1/2) × (du/u)  = (1/2)t + ln(u) + C  = (1/2)t + ln(cost + sint) + C  But, x = a sint   ....   x = 0, a  0 = a sint ⇒ sint = 0 ⇒ t = sin^(−1) 0 ⇒ t = 0  Again,  a = a sint ⇒ sint = 1 ⇒ t = sin^(−1) (1) ⇒ t = 90 ⇒ t = (Π/2)  [(1/2)t + ln(sint + cost) + C]_0 ^(Π/2)   = (1/2)((Π/2))+ln(sin(Π/2)+cos(Π/2))+C − [(1/2)(0)+ln(sin0+cos0) + C]  = (Π/4) + ln(1+0)+C−0−ln(0+1)−C  = (Π/4) + ln(1) + C − 0 − ln(1) − C  = (Π/4)     DONE !  Force × Distance = ISE
x=asintdx=acostdtdxx+a2x2=acostdtasint+a2a2sin2t=acostdtasint+a2a2(1cos2t)=acostdtasint+a2a2+a2cos2t=acostdtasint+a2cos2t=acostdtasint+acost=acostdta(sint+cost)=costdtsint+cost=12×2costdtsint+cost=12×(cost+cost)dtsint+cost=12×(cost+sint+costsint)dtsint+cost=12(cost+sint)dt+12(costsint)dtcost+sint=12×cost+sintcost+sintdt+12×costsintcost+sintdt=12dt+12×costsintcost+sintdtletu=cost+sint,dudt=sint+cost,du=(costsint)dt=12dt+12×duu=12t+ln(u)+C=12t+ln(cost+sint)+CBut,x=asint.x=0,a0=asintsint=0t=sin10t=0Again,a=asintsint=1t=sin1(1)t=90t=Π2[12t+ln(sint+cost)+C]0Π2=12(Π2)+ln(sinΠ2+cosΠ2)+C[12(0)+ln(sin0+cos0)+C]=Π4+ln(1+0)+C0ln(0+1)C=Π4+ln(1)+C0ln(1)C=Π4DONE!Force×Distance=ISE
Commented by sandy_suhendra last updated on 01/Jul/16
It′s a great answer.   But I think there is only a little mistake in typing  sin t = 1 ⇒ t = 90 (not 45) = (π/2)
Itsagreatanswer.ButIthinkthereisonlyalittlemistakeintypingsint=1t=90(not45)=π2
Commented by Tawakalitu. last updated on 01/Jul/16
corrected. 10Q
corrected.10Q

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