integrate over 0 to 1 sin inverse[x.rootover(1-x) -rootx.rootover(1-xsquare)]

Asked by Debasish Panigrahi | 14th Feb, 2012, 08:41: AM

Expert Answer:

Let root(x) = sinA , x = sinB
Put in the equation we get
sin-1{ sinB . root(1 - sin2A) - sinA . root(1-sin2B) } = sin-1 { sinB.cosA - cosB.sinA }
= sin-1 { sin(B-A) } = B - A = sin-1x - sin-1(rootx)
The above expression can be integrated using By Parts .

Answered by  | 14th Feb, 2012, 03:22: PM

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