Answer to Question #176984 in Calculus for Phyroe

Question

Integrals Giving Inverse Trigonometric Functions

∫(e^x dx)/√(1-e^(2x))

Answer 1

Put $e^x=u$, then $x=ln u, dx = \frac{du}{u}$. Also, $e^{2x}=(e^x)^2=u^2$. Hence,

$$\intop \frac{e^x dx}{\sqrt{1-e^{2x}}}= \intop \frac{u\cdot \frac{du}{u}}{\sqrt{1-u^2}}=\intop \frac{du}{\sqrt{1-u^2}}=$$

$$=sin^{-1}u +C=sin^{-1}e^x +C$$

Answer: $sin^{-1}e^x +C$