Question

Please help me get started on the following trigonometric substitution problem (click to see).

Answer 1

\[\int\limits_{}^{}e^x \sqrt{1-e ^{2x}}dx\]

Answer 2

How do I write the given integral as one of the following...
x=asin(theta)
x=atan(theta)
x=asec(theta)

Answer 3

Try getting rid of the exponential with a substiution u=...

Answer 4

ok ill try that

Answer 5

\[\int\limits_{}^{}u \sqrt{1-u^2}\frac{ du }{ e^x }\]

Answer 6

@xavier

Answer 7

or do i do this...
\[\int\limits_{}^{}e^x \sqrt{1-u^2}\frac{ du }{ e^x }\]

Answer 8

Don't turn that e^x in front of your radical into a u. Then it cancels nicely.
Yes like that

Answer 9

ok then...
\[e^x=\sin \theta\]
and i plug that back into the original integral?

Answer 10

Just work with the u's for now. Cancel out the e^x in front of the radical with the one under the du. Then you have a new integral which you can treat by itself forgetting all about the e^x. Then you will come back at the end and substitute back.

Answer 11

ok i'll show you my result when i get there

Answer 12

Good luck

Answer 13

but when i plug sin(theta) into the equation i am integrating with respect to theta and i don't have any u's left...how do i know where to plug the e^x back in?

Answer 14

\[\sqrt{1-\sin ^{2}\theta}\]
becomes
\[\sqrt{\cos ^{2}\theta}\]

Answer 15

and i quickly lose track of sin(theta)

Answer 16

Thats what you want to happen. Because when you do
u = sin(theta) you have du = cos(theta) d(theta). Give it a try and see how it cancels

Answer 17

ok i integrated on this is the result
\[\theta+\sin(2\theta)+C\]

Answer 18

and* this is the result

Answer 19

Something went wrong
integral of sqrt(1-u^2)du should have turned to integral of cos(theta) d(theta)
Check the substiutiton one more time

Answer 20

du=cos(theta) d(theta)

Answer 21

Ah sorry. You end up with the integral of cos^2(theta). Are you sure your answer isn't missing a factor of 1/2 along the way or a bit more. I'm assuming you used power reduction identity

Answer 22

w=2(theta)
(1/2)w+sin(w)+c = (theta)+sin(2theta)+c

Answer 23

and that is after
\[\int\limits_{}^{}\cos^2(\theta) d \theta=\frac{ 1 }{ 2 }\int\limits_{}^{}(1+\cos(2\theta)d \theta\]

Answer 24

Let me call theta x here for sake of less typing
So now you have 1/2x + 1/4sin(2x) + C.

Answer 25

it's not (1/2)sin(2x)?

Answer 26

But you have the 1/2 outside already

Answer 27

where did you get (1/4)

Answer 28

What is the integral of just cos(2x)?

Answer 29

oh ok i see now