If ∫exxsin−1⁡x1−x2+sin−1⁡x1−x23/2+x1−x2dx=g(x)+C, where C is the constant of integration, then g12 equals :

∵ddxxsin−1⁡x1−x2=sin−1⁡x1−x23/2+x1−x2⇒∫exxsin−1⁡x1−x2+sin−1⁡x1−x23/2+x1−x2dx=ex⋅xsin−1⁡x1−x2+c=g(x)+CNote : assuming g(x)=xexsin−1⁡x1−x2g(1/2)=e1/22⋅π6×23=π6e3Comment : In this question we will not get a unique function g(x), but in order to match the answer we will have to assume g(x)=xexsin−1⁡x1−x2.

If ∫exxsin−1⁡x1−x2+sin−1⁡x1−x23/2+x1−x2dx=g(x)+C, where C is the constant of integration, then g12 equals :

∵ddxxsin−1⁡x1−x2=sin−1⁡x1−x23/2+x1−x2⇒∫exxsin−1⁡x1−x2+sin−1⁡x1−x23/2+x1−x2dx=ex⋅xsin−1⁡x1−x2+c=g(x)+CNote : assuming g(x)=xexsin−1⁡x1−x2g(1/2)=e1/22⋅π6×23=π6e3Comment : In this question we will not get a unique function g(x), but in order to match the answer we will have to assume g(x)=xexsin−1⁡x1−x2.

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If $\int e^{x} (\frac{{xsin}^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{\sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} + \frac{x}{1 - x^{2}}) dx = g (x) + C$ , where C is the constant of integration, then $g (\frac{1}{2})$ equals :

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$\frac{π}{6} \sqrt{\frac{e}{2}}$

$\frac{π}{4} \sqrt{\frac{e}{3}}$

$\frac{π}{4} \sqrt{\frac{e}{2}}$

$\frac{π}{6} \sqrt{\frac{e}{3}}$

answer is C.

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Detailed Solution

$\begin{array}{l} ∵ \frac{d}{dx} (\frac{{xsin}^{- 1} x}{\sqrt{1 - x^{2}}}) = \frac{\sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} + \frac{x}{1 - x^{2}} \\ \Rightarrow \int e^{x} (\frac{{xsin}^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{\sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} + \frac{x}{1 - x^{2}}) dx \\ = e^{x} \cdot \frac{{xsin}^{- 1} x}{\sqrt{1 - x^{2}}} + c = g (x) + C \end{array}$
Note : assuming $g (x)= \frac{{xe}^{x} \sin^{- 1} x}{\sqrt{1 - x^{2}}}$
$g (1 / 2) = \frac{e^{1 / 2}}{2} \cdot \frac{\frac{π}{6} \times 2}{\sqrt{3}} = \frac{π}{6} \sqrt{\frac{e}{3}}$
Comment : In this question we will not get a unique function g(x), but in order to match the answer we will have to assume $g (x) = \frac{{xe}^{x} \sin^{- 1} x}{\sqrt{1 - x^{2}}}$ .