Question

Integration:

Answer 1

\[\int\limits_{}^{}\frac{ dx }{ e^{x} + e^{-x} }\]
I need help with the above. Maybe I can replace e*x = t. but then i get stuck...

Answer 2

looks pretty good @tomiko to me.

If

\[\Large u=e^x \]
then
\[\Large u^{-1}=\frac{1}{u} =e^{-x}\]
now I recommend you to differentiate:
\[\Large u=e^x \]

Answer 3

this kinda looks like 1/cosh or 1/sinh

Answer 4

possible @dan815, but if you carry up the substitution above you will end up with a pretty famous integral, namely:
\[\Large \int\frac{1}{u^2+1}du \]

Answer 5

@IrishBoy123 WHY would you make u = tanTHETA? i dont get it. any basis?

Answer 6

only because tan^2 + 1 = sec^2 - from Pythagoreas, so the bottom becomes sec^2
but also because d(tanø)dø = sec^2 ø from the substitution so you get a sec^2 on the top too. they cancel out very cleanly.

anytime you see squared terms, you can look to the trig identities such as 1 - cos^ 2 = sin^2 etc etc as a potential solution.

Answer 7

Note the following:

After substituing \(\bf e^{x} = u\), your integral looks like what @spacelimbus typed up. Now note that:\[\bf \frac{ d }{ dx }\tan^{-1}(u)=\frac{ 1 }{ 1+u^2 }\frac{ du }{ dx }\]

Answer 8

@tomiko

Answer 9

So the anti-derivative would be?....

@tomiko

Answer 10

the anti-deriv or integral is arctan(exp(x))

in that sheet i attached, you undo the 2 substitutions in the last 2 lines to show that this is the answer.

Answer 11

\[\large \int\limits \frac{1}{e^x+e^{-x}}dx\]\[\large u= e^x \]\[\large du= e^x dx \ \therefore \ dx=\frac{du}{e^x}=\frac{du}{u}\]\[\large \frac{1}{u}=\frac{1}{e^{-x}}\]\[\large \int\limits \frac{1}{u+\frac{1}{u}}\cdot \frac{du}{u}= \int\limits \frac{1}{\frac{u^2+1}{u}}\cdot \frac{du}{u}= \int\limits \frac{1}{u^2+1}du\]\[\large = \tan^{-1}(u)+c = \tan^{-1}(e^{x})+c\]

Answer 12

thanks @Jhannybean :)