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?

Thanks in advance

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.

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

Re: Integrate e^x(1-e^2x)^(1/2)dx

Hello, verasi!

Your substitution is faulty.

We have: .

Let

Substitute: .

Let

Substitute: .

. . . . . . . .

Back-substitute: .

Back-substitute: .

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

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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 :