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Evaluate the integrals by making appropriate $u$ -substitutions and applying the formulas reviewed in this section.$$\int \frac{e^{x}}{\sqrt{1-e^{2 x}}} d x$$

$\int \frac{e^{x}}{\sqrt{1-e^{2 x}}} d x=\arcsin \left(e^{x}\right)+C$

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

PRINCIPLES OF INTEGRAL EVALUATION

Section 1

An Overview of Integration Methods

Integrals

Integration

Integration Techniques

Trig Integrals

Trig Substitution

Campbell University

Baylor University

University of Michigan - Ann Arbor

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So here we have the integral of eat of the X over the square root of one minus eat of the two X. So to solve this, we're gonna use the use substitution where U equals e to the X, which means that d you is just either the axe DX. It means we can instead of the numerator e to the x dx, we'd just sub student d'you over the square root of one minus. You squared. Now we're gonna use a trigger substitution where instead of you we're gonna set you equal to sign of Z. Let's say which means that, um d u equals, um, the co sign of Z d Z And the reason we're doing this is ultimately we're going to use the identity one minus sine squared. X equals co sine squared X. If we get if we can use that identity, things become very easy. So instead of d u in the numerator we have co sign Z DZ over square root of one minus sine squared Z one minus sine squared Z We've already established its co sine squared of Z. Now we have co signs e TZ over the square root of co sine squared, which is just co signs E Now we see these cancel out. So all we're left with is the integral of DZ, which we know from the internal table is Jay Z plus the constant of integration. Now, um, we have to change this, z, um, back into you and then ultimately back into in terms of X. So we have you equal sine z, which means that Z equals inverse sign of you so this can become in verse sign of you plus c you we already know is e to the X. So all this becomes is inverse sign of e to the X plus the constant of integration.

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