$\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \mathrm{~d} x=\sin ^{-1}(f(x))+C, \quad$ (where $\quad C \quad$ is constant of integration.) then $f(2)$ has the value

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Aicharya · Accepted Answer

$\begin{aligned} & \int \frac{e^{x / 2}}{\sqrt{e^{-x}-e^x}} \mathrm{~d} x=\int \frac{e^x \mathrm{~d} x}{\sqrt{1-\left(e^xight)^2}}\left[	ext { Multiplying } N^r 	ext { and } D^r 	ext { by } e^{\frac{x}{2}} 	ext { ] }ight. \ & =\int \frac{\mathrm{d} t}{\sqrt{1-t^2}}=\sin ^{-1} t+C=\sin ^{-1}\left(e^xight)+C\left[	ext { let } e^x=tight] \ & \Rightarrow f(x)=e^x \ & \Rightarrow f(2)=e^2\end{aligned}$

$\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \mathrm{~d} x=\sin ^{-1}(f(x))+C, \quad$ (where $\quad C…

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