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(D) To evaluate $I = \int \sin^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \sin^{-1} x$ and $dv = dx$. The

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$x \sin^{-1} x + \sqrt{1 - x^2} + c$

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(D) To evaluate $I = \int \sin^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \sin^{-1} x$ and $dv = dx$. Then $du = \frac{1}{\sqrt{1 - x^2}} \, dx$ and $v = x$. Applying the formula: $I = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx$. To solve the integral $\int \frac{x}{\sqrt{1 - x^2}} \, dx$,let $t = 1 - x^2$,so $dt = -2x \, dx$,which means $x \, dx = -\frac{1}{2} dt$. Substituting this: $\int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot (2t^{1/2}) = -\sqrt{t} = -\sqrt{1 - x^2}$. Therefore,$I = x \sin^{-1} x - (-\sqrt{1 - x^2}) + c = x \sin^{-1} x + \sqrt{1 - x^2} + c$.

$\int \sin^{-1} x \, dx$ is equal to

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