Evaluate Trig Subs Integral w/ e^x = sin∅

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Homework Help Overview

The discussion revolves around evaluating an integral using trigonometric substitution, specifically involving the expression \( e^x = \sin \theta \). Participants are exploring the setup and manipulation of the integral \(\int e^x\,dx / \sqrt{1-e^{2x}}\).

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the substitution \( e^x = \sin \theta \) and its implications for the integral. There are attempts to express the differential \( dx \) in terms of \( d\theta \) and to simplify the integral accordingly. Some participants question the correctness of the expressions used, particularly regarding the square root and the form of the integral.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's attempts. There is recognition of errors in the setup, particularly concerning the square root and the integral's form. While some corrections have been suggested, there is no explicit consensus on the final approach or solution.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can provide or the methods they can use. The discussion reflects a collaborative effort to clarify the steps involved in the trigonometric substitution process.

whatlifeforme
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Homework Statement


Use a trigonometric substitution to evaluate the integral.


Homework Equations


\int e^x\,dx /\sqrt{1-e^2x}


The Attempt at a Solution


e^x = sin∅
x=lnsin∅
dx=cos∅/sin∅


\frac{sin∅*cos∅}{sin∅*\sqrt{1-(sin∅)^2}}





\int sin∅cos∅<br /> /<br /> sin∅(cos∅)\,d∅



\int \,d∅ = ∅

e^x = sin∅
∅ = arcsin(e^x)

answer:
∅ + C
arcsin(e^x) + c
 
Last edited:
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hi whatlifeforme! :smile:
whatlifeforme said:
sin∅/sqrt(1-(sin∅)^2)



\int sin∅cos∅<br /> /<br /> sin∅(cos∅)^2\,dx

nooo :redface:

you forgot the sqrt :wink:

(btw it's easier to say ex = sinθ, so exdx = cosθdθ)
 
tiny-tim said:
hi whatlifeforme! :smile:


nooo :redface:

you forgot the sqrt :wink:

(btw it's easier to say ex = sinθ, so exdx = cosθdθ)

so it should be

\int sin∅cos∅ <br /> / <br /> sin∅(cos∅)\,dx
 
whatlifeforme said:
so it should be

\int sin∅cos∅ <br /> / <br /> sin∅(cos∅)\,dx
thus,\int 1\,d∅

e^x = sin∅
∅ = arcsin(e^x)

answer: arcsin(e^x) + c
 
Last edited:
yup! :biggrin:

(except of course that integral should have been ∫ 1 dθ :wink:)
 
tiny-tim said:
yup! :biggrin:

(except of course that integral should have been ∫ 1 dθ :wink:)

ooopss. sry fixed.
 

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