The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Integration by Substitution Hint

Hey,

Can anyone give me a hint on how to integrate

∫ex1−e2x \int \frac{e^x}{\sqrt{1-e^{2x}}} . So far I've only come up with the fact that exe^x and 1−e2x\sqrt{1-e^{2x}} are the legs of a right angled triangle with hypotenuse of length 1...
Reply 1
Avatar for Notnek
Notnek
21
Original post by FXLander
Hey,

Can anyone give me a hint on how to integrate

∫ex1−e2x \int \frac{e^x}{\sqrt{1-e^{2x}}} . So far I've only come up with the fact that exe^x and 1−e2x\sqrt{1-e^{2x}} are the legs of a right angled triangle with hypotenuse of length 1...

Have you tried any substitutions?
i think a trig substitution for ex would work...
Reply 3
Avatar for FXLander
FXLander
OP
10
Original post by Notnek
Have you tried any substitutions?


I tried substituting tan⁡θ=ex1−e2x\tan \theta = \frac{e^x}{\sqrt{1-e^{2x}}} since I drew the right triangle I was referring to, but I am not sure how to get rid of the dx when I do this...
Reply 4
Avatar for Notnek
Notnek
21
Original post by FXLander
I tried substituting tan⁡θ=ex1−e2x\tan \theta = \frac{e^x}{\sqrt{1-e^{2x}}} since I drew the right triangle I was referring to, but I am not sure how to get rid of the dx when I do this...

You're making it more complicated than it needs to be. Try the substitution that the bear gave you. With practice this will feel like an "obvious" substitution.

EDIT : i.e. I mean u=exu=e^x (this sub will lead to a standard/formula book integral)
(edited 6 years ago)
Original post by FXLander
I tried substituting tan⁡θ=ex1−e2x\tan \theta = \frac{e^x}{\sqrt{1-e^{2x}}} since I drew the right triangle I was referring to, but I am not sure how to get rid of the dx when I do this...


use a different trig function for e^x. consider trig identities.
Reply 6
Avatar for FXLander
FXLander
OP
10
Original post by Notnek
You're making it more complicated than it needs to be. Try the substitution that the bear gave you. With practice this will feel like an "obvious" substitution.

EDIT : i.e. I mean u=exu=e^x


After doing the substitution I got ∫du1−u2 \int \frac{du}{\sqrt{1-u^2}} which I then evaluated to 1−e2xex\frac{\sqrt{1-e^{2x}}}{e^x} , however I am not sure whether this is correct...
Reply 7
Avatar for simon0
simon0
13
Original post by FXLander
After doing the substitution I got ∫du1−u2 \int \frac{du}{\sqrt{1-u^2}} which I then evaluated to 1−e2xex\frac{\sqrt{1-e^{2x}}}{e^x} , however I am not sure whether this is correct...


For:

∫11−u2 du, \displaystyle \int \dfrac{1}{ \sqrt{1-u^{2}} } \, du,

look up a table of inverse trigonometric derrivatives.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you