int {{e^x}left[ {{{sin }^{ - 1}}frac{x}{a} + | vedclass

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(C) We use the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Let $f(x) = \sin^{-1}(\frac{x}{a})$. Then,the derivative $f'(x)

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${e^x}{\sin ^{ - 1}}\frac{x}{a} + c$

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(C) We use the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Let $f(x) = \sin^{-1}(\frac{x}{a})$. Then,the derivative $f'(x) = \frac{d}{dx}(\sin^{-1}(\frac{x}{a})) = \frac{1}{\sqrt{1 - (x/a)^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \cdot \frac{1}{a} = \frac{a}{\sqrt{a^2 - x^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{a^2 - x^2}}$. Since the integrand is of the form $e^x [f(x) + f'(x)]$,the integral is $e^x f(x) + c$. Therefore,$\int e^x [\sin^{-1}(\frac{x}{a}) + \frac{1}{\sqrt{a^2 - x^2}}] dx = e^x \sin^{-1}(\frac{x}{a}) + c$.

$\int {{e^x}\left[ {{{\sin }^{ - 1}}\frac{x}{a} + \frac{1}{{\sqrt {{a^2} - {x^2}} }}} \right]dx = }$

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