A) \[{{e}^{{{\sin }^{-1}}x}}({{\sin }^{-1}}x-1)+c\]
B) \[{{e}^{{{\sin }^{-1}}x}}+c\]
C) \[{{e}^{{{({{\sin }^{-1}}x)}^{2}}}}+c\]
D) \[{{e}^{2{{\sin }^{-1}}x}}+c\]
E) \[{{e}^{{{\sin }^{-1}}x}}{{\sin }^{-1}}x+c\]
Correct Answer: A
Solution :
Given that, \[f(x)=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\]and \[g(x)={{e}^{{{\sin }^{-1}}x}}\] \[\therefore \] \[\int{f(x)}g(x)dx=\int{\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}{{e}^{{{\sin }^{-1}}x}}dx\] Let\[{{\sin }^{-1}}x=t\]and, \[\frac{1}{\sqrt{1-{{x}^{2}}}}dx=dt\] \[\Rightarrow \]\[\int{f(x)g(x)}dx=\int{t{{e}^{t}}}dt=t{{e}^{t}}-{{e}^{t}}+c\] \[={{e}^{{{\sin }^{-1}}x}}({{\sin }^{-1}}x-1)+c\]
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