If f(x) = frac{sin^{-1} x}{sqrt{1-x^2}} and g | vedclass

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(A) Let $I = \int f(x)g(x) dx = \int \frac{\sin^{-1} x}{\sqrt{1-x^2}} e^{\sin^{-1} x} dx$.Substitute $u = \sin^{-1} x$, then $du = \frac{1}{\sqrt{1-x^

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int f(x)g(x) dx = \int \frac{\sin^{-1} x}{\sqrt{1-x^2}} e^{\sin^{-1} x} dx$.Substitute $u = \sin^{-1} x$, then $du = \frac{1}{\sqrt{1-x^2}} dx$.The integral becomes $I = \int u e^u du$.Using integration by parts $\int u e^u du = u e^u - \int e^u du = u e^u - e^u + c$.Substituting back $u = \sin^{-1} x$, we get $I = e^{\sin^{-1} x}(\sin^{-1} x - 1) + c$.

If $f(x) = \frac{\sin^{-1} x}{\sqrt{1-x^2}}$ and $g(x) = e^{\sin^{-1} x}$, then the value of $\int f(x)g(x) dx = \dots$

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