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(D) To evaluate the integral $I = \int \cos^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = \cos^

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

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(D) To evaluate the integral $I = \int \cos^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = \cos^{-1} x$ and $dv = dx$.Then $du = -\frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.Applying the formula: $I = x \cos^{-1} x - \int x \left( -\frac{1}{\sqrt{1-x^2}} \right) \, dx$.$I = x \cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} \, dx$.To solve $\int \frac{x}{\sqrt{1-x^2}} \, dx$,let $t = 1-x^2$,so $dt = -2x \, dx$ or $x \, dx = -\frac{1}{2} \, dt$.The integral becomes $\int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) = -\sqrt{t} = -\sqrt{1-x^2}$.Therefore,$I = x \cos^{-1} x - \sqrt{1-x^2} + c$.

$\int \cos^{-1} x \, dx =$

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