If $\int e^x \left(\frac{x \sin^{-1}x}{\sqrt{1-x^2}} + \frac{\sin^{-1}x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}\right) dx = g(x) + C$, where C is the constant of integration, then $g(\frac{1}{2})$ equals:
- (1) $\frac{\pi}{6}\sqrt{\frac{e}{2}}$
- (2) $\frac{\pi}{4}\sqrt{\frac{e}{2}}$
- (3) $\frac{\pi}{6}\sqrt{\frac{e}{3}}$
- (4) $\frac{\pi}{4}\sqrt{\frac{e}{3}}$
Solution:
Let $f(x) = \frac{x \sin^{-1}x}{\sqrt{1-x^2}}$.
Differentiating $f(x)$:
$$ f'(x) = \frac{\sqrt{1-x^2}(\sin^{-1}x + \frac{x}{\sqrt{1-x^2}}) – x\sin^{-1}x(\frac{-2x}{2\sqrt{1-x^2}})}{1-x^2} $$
$$ f'(x) = \frac{\sin^{-1}x}{\sqrt{1-x^2}} + \frac{x}{1-x^2} + \frac{x^2 \sin^{-1}x}{(1-x^2)^{3/2}} $$
Simplifying $\frac{\sin^{-1}x}{\sqrt{1-x^2}} + \frac{x^2 \sin^{-1}x}{(1-x^2)^{3/2}}$:
$$ = \frac{(1-x^2)\sin^{-1}x + x^2\sin^{-1}x}{(1-x^2)^{3/2}} = \frac{\sin^{-1}x}{(1-x^2)^{3/2}} $$
Thus, the term inside the bracket is $f(x) + f'(x)$.
Integral is $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
So, $g(x) = e^x \frac{x \sin^{-1}x}{\sqrt{1-x^2}}$.
Calculate $g(\frac{1}{2})$:
$$ g(1/2) = e^{1/2} \frac{\frac{1}{2} \sin^{-1}(1/2)}{\sqrt{1-(1/4)}} = \sqrt{e} \frac{\frac{1}{2} \cdot \frac{\pi}{6}}{\frac{\sqrt{3}}{2}} $$
$$ = \sqrt{e} \frac{\pi}{12} \cdot \frac{2}{\sqrt{3}} = \frac{\pi}{6} \frac{\sqrt{e}}{\sqrt{3}} = \frac{\pi}{6}\sqrt{\frac{e}{3}} $$
Ans. (3)