# Long Integration Question

Integrate $$2\pi \int e^{-x}\sqrt{1+e^{-2x}} .dx$$

By looking at the question i knew it would lead to alot of calculations and working out but i've tried sooo many different methods and just end up in a mess.

-

Setting $t=\exp(-x)$ as an substitution, so you have $$dt=-\exp(-x)dx\longrightarrow\text{e}^{-x}dx=-dt$$ so we can convert the original integral to: $$-2\pi\int\sqrt{1+\left(\text{e}^{-x}\right)^2}dt\to -2\pi\int\sqrt{1+t^2}dt\\=-2\pi\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}\sinh^{-1}t+C, \;\;t=\text{e}^{-x}$$

We already know that $\int \sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\sinh^{-1}\frac{x}{a}+C\;^*$ which can be proved by using a nice substitution $x=a\sinh(\theta)$ or using http://planetmath.org/?method=l2h&id=9681&op=getobj&from=objects.Try to verify above formula$^*$ for yourself and then use it for your integral . $\ddot\smile$

-
(+1) nice hint. – Mhenni Benghorbal Jan 3 '13 at 17:40
@MhenniBenghorbal: Thanks for your support. – Babak S. Jan 3 '13 at 17:43
i've understood the first few steps and how you've come to the conclusion of $$-2\pi\int\sqrt{1+t^2}dt$$ however after that i dont really know what i'm doing – jill Jan 3 '13 at 18:04
@jill: See my recent edit and tell me if you need any help. – Babak S. Jan 3 '13 at 18:21
@BabakSorouh ohhh i made a simple calculation mistake... Thankyouuu :D! – jill Jan 3 '13 at 18:22

Did you try to put $u=e^{-x}$ ? It seems that it gives you a nice formula : $-2\pi \int \sqrt{1+u^2}du$

-