Q.1 (vi) If f(x) = fracsin⁻¹x√(1-x²) and g(x) = e^sin⁻¹x then ∫ f(x)· g(x) dx = _____. A. e^sin⁻¹x(1 - sin⁻¹x) + c B. e^sin⁻¹x(sin⁻¹x - 1) + c C. e^sin⁻¹x(sin⁻¹x + 1) + c D. e^cos⁻¹x(cos⁻¹x - 1) + c

Answer: B. e^sin⁻¹x(sin⁻¹x - 1) + c Answer: (b) e^sin⁻¹x(sin⁻¹x - 1) + c Put sin⁻¹x = t , so dx/√(1-x²) = dt . The integral becomes ∫ t e^t dt = te^t - ∫ e^t dt = e^t(t - 1) + c = e^sin⁻¹x(sin⁻¹x - 1) + c .

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

Q.1 (vi) If \(f(x) = \frac{\sin^{-1}x}{\sqrt{1-x^2}}\) and \(g(x) = e^{\sin^{-1}x}\) then \(\int f(x)\cdot g(x)\,dx =\) _____.

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

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

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

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

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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

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

Put \(\sin^{-1}x = t\), so \(\frac{dx}{\sqrt{1-x^2}} = dt\). The integral becomes \(\int t e^t\,dt = te^t - \int e^t\,dt = e^t(t - 1) + c = e^{\sin^{-1}x}(\sin^{-1}x - 1) + c\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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

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