If int frac{e^{frac{x}{2}}}{sqrt{e^{-x}-e^x}} | vedclass

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(B) Let $I = \int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx$. Multiply the numerator and denominator by $e^{\frac{x}{2}}$: $I = \int \frac{e^{\f

Vedclass · Accepted Answer

$e^2$

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(B) Let $I = \int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx$. Multiply the numerator and denominator by $e^{\frac{x}{2}}$: $I = \int \frac{e^{\frac{x}{2}} \cdot e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x} \cdot e^{\frac{x}{2}}} \, dx = \int \frac{e^x}{\sqrt{e^{-x} \cdot e^{\frac{x}{2}} - e^x \cdot e^{\frac{x}{2}}}} \, dx$. Wait,let us simplify the denominator differently: $I = \int \frac{e^{\frac{x}{2}}}{\sqrt{\frac{1}{e^x} - e^x}} \, dx = \int \frac{e^{\frac{x}{2}}}{\sqrt{\frac{1 - e^{2x}}{e^x}}} \, dx = \int \frac{e^{\frac{x}{2}} \cdot \sqrt{e^x}}{\sqrt{1 - e^{2x}}} \, dx$. Since $\sqrt{e^x} = e^{\frac{x}{2}}$,we have: $I = \int \frac{e^{\frac{x}{2}} \cdot e^{\frac{x}{2}}}{\sqrt{1 - (e^x)^2}} \, dx = \int \frac{e^x}{\sqrt{1 - (e^x)^2}} \, dx$. Let $t = e^x$,then $dt = e^x \, dx$. $I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + C = \sin^{-1}(e^x) + C$. Comparing this with $\sin^{-1}(f(x)) + C$,we get $f(x) = e^x$. Therefore,$f(2) = e^2$.

If $\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx = \sin^{-1}(f(x)) + C$,(where $C$ is the constant of integration),then $f(2)$ has the value:

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