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$\begin{array}{l} \int e^{\cos ^{-1} x}\left[\frac{x-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right] d x=\\\end{array}$ (A) $-e^{\sin ^{-1} x}+c$ (B) $-x e^{\cos ^{-1} x}+c$ (C) $-x e^{\sin ^{-1} x}+c$ (D) $-e^{\cos ^{-1} x}+c$
$\begin{array}{l} \int e^{\cos ^{-1} x}\left[\frac{x-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right] d x=\\\end{array}$ (A) $-e^{\sin ^{-1} x}+c$ (B) $-x e^{\cos ^{-1} x}+c$ (C) $-x e^{\sin ^{-1} x}+c$ (D) $-e^{\cos ^{-1} x}+c$
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$\begin{array}{l} \int e^{\cos ^{-1} x}\left[\frac{x-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right] d x=\\\end{array}$ (A) $-e^{\sin ^{-1} x}+c$ (B) $-x e^{\cos ^{-1} x}+c$ (C) $-x e^{\sin ^{-1} x}+c$ (D) $-e^{\cos ^{-1} x}+c$

The correct option is option (B) $-x e^{\cos ^{-1} x}+c$

$\int \frac{\left(x-\sqrt{1-x^{2}}\right) e^{\ ccos (x)}}{\sqrt{1-x^{2}}} \mathrm{~d} x$

$=-\int \frac{\left(\sqrt{1-x^{2}}-x\right) e^{\ ccos (x)}}{\sqrt{1-x^{2}}} \mathrm{~d} x$

$\begin{array}{c}
\int \frac{\left(\sqrt{1-x^{2}}-x\right) e^{\ ccos (x)}}{\sqrt{1-x^{2}}} d x \\
\\
=\int\left(e^{\ ccos (x)}-\frac{x e^{\ ccos (x)}}{\sqrt{1-x^{2}}}\right) \mathrm{d} x
\end{array}$

$=\int e^{\ ccos (z)} \mathrm{d} x-\int \frac{x e^{\ ccos (x)}}{\sqrt{1-x^{2}}} \mathrm{~d} x$

$\int \frac{x e^{\ ccos (x)}}{\sqrt{1-x^{2}}} \mathrm{~d} x: \int \mathrm{fg}^{\prime}=\mathbf{f} g-\int \mathrm{f}^{\prime} \mathrm{g}$

$\mathrm{f}=\boldsymbol{x}, \quad \mathrm{g}^{\prime}=\frac{e^{\operatorname{ccos}(\mathbf{z})}}{\sqrt{1-x^{2}}}$

$f^{\prime}=1, \quad g=-e^{\ ccos (x)}$

$\left.=\mathcal{L} e^{ cos (x)}+\int-\mathbf{e}^{\operatorname{cos}(x)} \mathrm{d} x+\int \mathrm{e}^{\arccos (x)} \mathrm{d} x\right)$

$\int \mathrm{e}^{\arccos (x)} \mathrm{d} x$

$=\boldsymbol{x}^{\operatorname{cos}(x)}$

$\begin{array}{c}
-\int \frac{\left(\sqrt{1-x^{2}}-x\right) e^{\ cos (x)}}{\sqrt{1-x^{2}}} d x \\
=-x e^{2 \operatorname{ cos}(x)}
\end{array}$

$\begin{array}{c}
\int \frac{\left(x-\sqrt{1-x^{2}}\right) e^{\ cos (x)}}{\sqrt{1-x^{2}}} d x \\
=-x e^{-1}{\ cos (x)}+C
\end{array}$

i

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