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Intigration of root sinX
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Intigration of root sinX?

∫ √sinx dx ___ eq (1)here we put sin x = y^2so that x = sin^-1 y^2diff wrt to ydx/dy = 1/√1-{(y^2)^2}dx/dy = 1/ √1-(y^4) ____ eq(2)sub (2)in(1)I= ∫ ( √y^2) × 1/√(1-y^4) dyI = ∫ y/√(1-y^4) dy ____eq(3)let y^2 = tso 2y dy = dtand y dy = dt/2therefore eq (3) bcumsI = ∫ dt/ (2√1-t^2)I = 1/2 ∫ dt/√1-t^2 _____eq(4)let t = sin zso dt/dz = cos zdt = cosz dztherefore (4) bcmsI = 1/2 ∫ cos z dz / 1-sin^2zI = 1/2 ∫ cos z dz / cos zI =1/2 ∫ dzI= 1/2 z+ cI= 1/2 sin^-1 t + cI = 1/2 sin^-1 y^2 + cI = 1/2 sin^-1(sin x) + cI = 1/2 x+c

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