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

[whatlifeforme]

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

 

[tiny-tim]

hi whatlifeforme!

sin∅/sqrt(1-(sin∅)^2)



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

nooo …

you forgot the sqrt ​

(btw it's easier to say e^x = sinθ, so e^xdx = cosθdθ)

 

[whatlifeforme]

hi whatlifeforme!


nooo …

you forgot the sqrt ​

(btw it's easier to say e^x = sinθ, so e^xdx = cosθdθ)

so it should be

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

 

[whatlifeforme]

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

 

[tiny-tim]

yup!

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

 

[whatlifeforme]

yup!

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

ooopss. sry fixed.