int e^{cos ^{-1} x} left[ frac{x-sqrt{1-x^{2} | vedclass

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(B) Let $\cos ^{-1} x = t$. Then $x = \cos t$ and $dx = -\sin t \ dt$. Also,$\frac{dx}{\sqrt{1-x^2}} = -dt$. Substituting these into the integral: $I

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$-x e^{\cos ^{-1} x} + c$

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(B) Let $\cos ^{-1} x = t$. Then $x = \cos t$ and $dx = -\sin t \ dt$. Also,$\frac{dx}{\sqrt{1-x^2}} = -dt$. Substituting these into the integral: $I = \int e^t \left[ \frac{\cos t - \sin t}{\sin t} \right] (-dt) = \int e^t \left[ \frac{\sin t - \cos t}{\sin t} \right] dt$ Wait,let us simplify the expression directly: $I = \int e^{\cos ^{-1} x} \left( \frac{x}{\sqrt{1-x^2}} - 1 \right) dx$. Let $t = \cos ^{-1} x$,so $x = \cos t$ and $dx = -\sin t \ dt$. $I = \int e^t \left( \frac{\cos t}{\sin t} - 1 \right) (-\sin t \ dt) = \int e^t (\sin t - \cos t) dt$. Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$,we get: $I = e^t \sin t + c$. Since $t = \cos ^{-1} x$,$\sin t = \sqrt{1 - \cos^2 t} = \sqrt{1 - x^2}$. This does not match the options. Let us re-evaluate: $I = \int e^t (\sin t - \cos t) dt = -\int e^t (\cos t - \sin t) dt = -e^t \cos t + c$. Substituting back $t = \cos ^{-1} x$: $I = -e^{\cos ^{-1} x} \cdot x + c = -x e^{\cos ^{-1} x} + c$.

$\int e^{\cos ^{-1} x} \left[ \frac{x-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}} \right] dx =$

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