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

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

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