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You are watching: Sin−1(x) dx

This is a solution to the question, which I discovered on the web. The just bit I"m not certain of, is the transition from the 3rd last line to second last line (highlighted in bold). I would certainly have fixed ?(1/2)u^(-1/2) du as -1/2 (sin^(-1)x), but I"m assuming that when du is characterized, integration is ignored and we simply extract the (displaystyle sqrtu). Can anyone aid clarify this with me?Many kind of thanks.Q. Integprice sin^(-1) x dxAns.: combine by partslet u = arcsinx ; dv = dxdu = dx/?(1-x²) ; v = xx arcsinx - ?x/?(1-x²) dxlet u = 1-x²du = -2x dxx dx = (-1/2)dux arcsinx - ?(-1/2)du / ?(u)x arcsinx + ?(1/2)u^(-1/2) dux arcsinx + u^(1/2)x arcsinx + ?(1-x²) + C
I changed the answer a little bit for you. I think currently you will understand also let u = arcsinx ; dv = dxdu = dx/?(1-x²) ; v = xx arcsinx - ?x/?(1-x²) dxlet t = 1-x²dt = -2x dxx dx = (-1/2)dtx arcsinx - ?(-1/2)dt / ?(t)x arcsinx + ?(1/2)t^(-1/2) dtx arcsinx + t^(1/2)x arcsinx + ?(1-x²) + C
Apologies, yet I"m not seeing much distinction from the answer gave and what you"re showing right here, i.e. respecifying "let u = 1-x²" with "let t = 1-x²"? My question still stands. What is occuring in between "x arcsinx + ?(1/2)u^(-1/2) du" & "x arcsinx + u^(1/2)"? What integration has actually taken place? (displaystyle xarcsin(x),+,frac12int u^-1/2,du,=,xarcsin(x),+,frac12,cdot,fracu^-frac12,+,1-frac12,+,1,+,C,=,xarcsin(x),+,u^frac12,+,C)
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