# Integrate e^x(1-e^2x)^(1/2)dx

• Mar 2nd 2013, 10:25 PM
verasi
Integrate e^x(1-e^2x)^(1/2)dx
Hi, I was wondering if you guys could check my work for indefinite integral of (e^x)(sqrt{1-e^2x)

Let u=e^x

The integral becomes:

integral of sqrt(1-u^2)

Then I do trigonometric substitution where I let u=sinx, and I get:

1/2sin^-1(x) + (x/2)*(1 - x^2)^1/2 + c.

However, the answer in my textbook is -1/3(1-e^(2x))^(3/2).

I was just wondering what I did wrong for the question.

If my complete process is wrong, what should I have done to solve the question?

• Mar 2nd 2013, 11:51 PM
JJacquelin
Re: Integrate e^x(1-e^2x)^(1/2)dx
Hi !
there is a mishmash in your answer because the same symbol x is used for two different variables :
u=e^x
u=sinx
leading to e^x=sinx which is false.
• Mar 3rd 2013, 02:18 AM
ibdutt
Re: Integrate e^x(1-e^2x)^(1/2)dx
You are just right put e^x=u that gives e^x dx = du
the question becomes integral of (1-u^2)^1/2 du and that is a standard for for which the integral is given by
(a^2-x^2)^1/2 = x/2 (a^2-x^2)^1/2 + 1/2 a^2 sin^-1 (x/a) + C
• Mar 3rd 2013, 07:48 AM
Soroban
Re: Integrate e^x(1-e^2x)^(1/2)dx
Hello, verasi!

Quote:

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

We have: .$\displaystyle \int \sqrt{1-e^{2x}}\,(e^x\,dx)$

Let $\displaystyle u = e^x \quad\Rightarrow\quad du = e^x\,dx$

Substitute: .$\displaystyle \int \sqrt{1-u^2}\,du$

Let $\displaystyle u = \sin\theta \quad\Rightarrow\quad du = \cos\theta\,d\theta$

Substitute: .$\displaystyle \int \sqrt{1-\sin^2\!\theta}\,\cos\theta\,d\theta \;=\;\int\cos^2\!\theta\,d\theta \;=\;\tfrac{1}{2}\int(1 + \cos2\theta)\,d\theta$

. . . . . . . . $\displaystyle =\;\tfrac{1}{2}\left(\theta + \tfrac{1}{2}\sin2\theta\right) + C \;=\;\tfrac{1}{2}(\theta + \sin\theta\cos\theta) + C$

Back-substitute: .$\displaystyle \tfrac{1}{2}\left(\arcsin u + u\sqrt{1-u^2}\right) + C$

Back-substitute: .$\displaystyle \tfrac{1}{2}\left(\arcsin e^x + e^x\sqrt{1-e^{2x}}\right) + C$
• Mar 3rd 2013, 03:28 PM
verasi
Re: Integrate e^x(1-e^2x)^(1/2)dx
Thanks for the reply, I guess my initial answer was right then. Apparently the textbook has a different answer. It's -1/3(1-e^(2x))^(3/2).

I'm guessing the textbook used a different method then (integrate by part maybe? I didn't try it seemed way too complicated).

T.T wasted 30min on this question trying to figure out what I did wrong
• Mar 3rd 2013, 10:14 PM
JJacquelin
Re: Integrate e^x(1-e^2x)^(1/2)dx
Quote:

Originally Posted by verasi
Thanks for the reply, I guess my initial answer was right then. Apparently the textbook has a different answer. It's -1/3(1-e^(2x))^(3/2).

I'm guessing the textbook used a different method then (integrate by part maybe? I didn't try it seemed way too complicated).

T.T wasted 30min on this question trying to figure out what I did wrong

Hi!

Fortunately, any different method leads to the same result ! The results are different if there is a mistake somewhere (in calculus or if the integrals are not the same) See attachment :