int frac{e^x , dx}{sqrt{1 - e^{2x}}} = | vedclass

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(B) Let $e^x = t$. Then,differentiating both sides with respect to $x$,we get $e^x \, dx = dt$. Substituting these into the integral,we have: $\int \f

Vedclass · Accepted Answer

$-\cos^{-1}(e^x) + c$

Vedclass · Answer

(B) Let $e^x = t$. Then,differentiating both sides with respect to $x$,we get $e^x \, dx = dt$. Substituting these into the integral,we have: $\int \frac{dt}{\sqrt{1 - t^2}}$. We know that $\int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$ or $-\cos^{-1}(t) + c$. Since the options provided use the cosine inverse function,we use the identity $\int \frac{dt}{\sqrt{1 - t^2}} = -\cos^{-1}(t) + c$. Substituting $t = e^x$ back,we get $-\cos^{-1}(e^x) + c$.

$\int \frac{e^x \, dx}{\sqrt{1 - e^{2x}}} = $

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