Find the following indefinite integral. Express your answer in exact form. Simplify.?

e^(x) * dx / sqrt(4 - e^(2x))

4 - e^(2x) = 4 - 4 * sin(t)^2
e^(2x) = 4 * sin(t)^2
e^(x) = 2 * sin(t)
e^(x) * dx = 2 * cos(t) * dt

e^(x) * dx / sqrt(4 - e^(2x)) =>
2 * cos(t) * dt / sqrt(4 - 4 * sin(t)^2) =>
2 * cos(t) * dt / sqrt(4 * cos(t)^2) =>
2 * cos(t) * dt / (2 * cos(t)) =>
dt

Integrate

t + C

e^(x) = 2 * sin(t)
(1/2) * e^(x) = sin(t)
t = arcsin((1/2) * e^(x))

arcsin((1/2) * e^(x)) + C

From 0 to ln(2^(1/2))

arcsin((1/2) * e^(ln(2^(1/2))) - arcsin((1/2) * e^(0)) =>
arcsin((1/2) * 2^(1/2)) - arcsin((1/2) * 1) = >
pi/4 - pi/6 =>
3pi/12 - 2pi/12 =>
pi/12...Show more

int_[-sqrt(2)]^(0) (e^x)/sqrt[4 - (e^(2x))] dx

u = e^x; du = e^x dx; a = e^(-sqrt(2)); b = 1

int_[e^(-sqrt(2))]^(1) 1/sqrt(4 - (u^2)) du

u = 2*sin(t); du = 2*cos(t) dt; a = arcsin[e^(-sqrt(2))/2]; b = pi/6

int_{arcsin[e^(-sqrt(2))/2]}^(pi/6) [2*cos(t)/sqrt(4 - 4*sin^2(t))] dt

int_{arcsin[e^(-sqrt(2))/2]}^(pi/6) [2*cos(t)/(2*cos(t))] dt

int_{arcsin[e^(-sqrt(2))/2]}^(pi/6) dt

http://www.wolframalpha.com/input/?i=int_%7Barcsin%5Be%5E(-sqrt(2))%2F2%5D%7D%5E(pi%2F6)+dt...Show more