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in Indefinite Integral by (32.3k points)

If f(x) = \(\frac{sin^{-1}x}{\sqrt{1 - x^2}},\) \(g(x) = e^{sin^{-1}}x,\) then ∫ f(x) . g(x) . dx =

f(x) = (sin-1x)/(√1 - x2), g(x) = esin-1x, then ∫ f(x) . g(x) . dx =

(a) \(e^{sin^{-1}}x . (sin^{-1}x - 1) + c\)

(b) \(e^{sin^{-1}}x . (1 - sin^{-1}x) + c\)

(c) \(e^{sin^{-1}}x . (sin^{-1}x + 1) + c\)

(d) \(e^{sin^{-1}}x . (sin^{-1}X - 1) + c\)

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1 Answer

+1 vote
by (32.3k points)

Correct answer is (a) \(e^{sin^{-1}}x . (sin^{-1}x - 1) + c\)

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