Question Number 6519 by benny last updated on 30/Jun/16

Commented by Tawakalitu. last updated on 30/Jun/16

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](https://www.tinkutara.com/question/Q6526.png)
Commented by sandy_suhendra last updated on 01/Jul/16

Commented by Tawakalitu. last updated on 01/Jul/16
