Integration (1 Viewer)

asb_92 · Jun 3, 2009

hey ppl!

how do u integrate:

1/ (e^x + e^-x)

ie. (qu 9 in cambridge pg. 175 further ex 5)

thanks heaps

Drongoski · Jun 3, 2009

asb_92 · Jun 3, 2009

oh ummz okay lolz
thanks heaps

rocktheyard · Jun 3, 2009

I'm confused...

Wouldn't the answer

0.5 $ln(1+e^{2x})$

also work?

Drongoski · Jun 3, 2009

NO. Don't forget also e^2x = e^x . e^x

$\100dpi \int \frac{2e^{2x}}{1+e^{2x}}\ dx\ \ =\ \ ln( 1+e^{2x} )\ +\ C$

rocktheyard · Jun 3, 2009

Drongoski said:
NO. Don't forget also e^2x = e^x . e^x

$\100dpi \int \frac{2e^{2x}}{1+e^{2x}}\ dx\ \ =\ \ ln( 1+e^{2x} )\ +\ C$

Oh, right.

gurmies · Jun 3, 2009

rocktheyard said:
Oh, right.

Made the same mistake the first time I did this question

kwabon · Jun 3, 2009

Drongoski said:
$\100dpi \int \frac{1}{e^{x}+e^{-x}}\ dx\ \ =\ \ \int \frac{e^{x}}{1+e^{2x} }\ dx\ \ =\ \ \int \frac{1}{1+(e^{x})^{2}}\ d\ e^{x}\ \ =\ \ tan^{-1}e^{x}\ +\ C$

could you please explain to me, what did you just do there??? how did you get rid of the e^x in the numerator

sorry for the inconvenience

azureus88 · Jun 3, 2009

kwabon said:
could you please explain to me, what did you just do there??? how did you get rid of the e^x in the numerator

sorry for the inconvenience

you probably missed the d(e^x) part after it.

kwabon · Jun 3, 2009

azureus88 said:
you probably missed the d(e^x) part after it.

lol, thats the part i dont get :burn:

Drongoski · Jun 3, 2009

kwabon

$\100dpi \frac{\mathrm{d} e^{x}}{\mathrm{d} x}\ =\ e^{x}\ \Rightarrow \ e^{x}dx \ =\ de^{x}\\ \\ Similarly\ \ \frac{1}{2\sqrt{x}}dx\ =\ d\sqrt{x}\\ \\ \int \frac{1}{1+x^{2}}\ dx\ \ =\ \ tan^{-1}x\ +\ C\\ \\ \ \Rightarrow \ \ \int \frac{1}{1+elephant^{2}}\ d\ elephant\ =\ tan^{-1}(elephant)\ +\ C\\ \\ \Rightarrow \ \ \int \frac{1}{2\sqrt{x}(1+x)}\ dx\ \ =\ \ \int \frac{1}{1+(\sqrt{x})^{2}}\ d\sqrt{x}\ \ = \ \ tan^{-1}\sqrt{x}\ +\ C$

kwabon · Jun 4, 2009

^^ thats a pretty cool way, never learnt it that way, thanks yet again drongoski!!

riseek · Jun 4, 2009

hopefully i ll remeber that for future reference

hermine · Jun 7, 2009

kwabon said:
could you please explain to me, what did you just do there??? how did you get rid of the e^x in the numerator

sorry for the inconvenience

another way of doing that just by substitution where u=e^x, du=e^x dx

sub that in and you get S dx/(1+u^2) which is the tan^-1 standard integral

Integration (1 Viewer)

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