Question

\[ \int_{-\infty}^{0}\frac{1}{cosh(x))}dx \]

Answer 1

\[\int\limits_{-\infty}^{0}\frac{1}{\cosh(x))}dx\]

Answer 2

1/cosh = cosech

Answer 3

Use the temporary variable t = e^x then it works smoothly

Answer 4

\[\int\limits_{0}^{1}\frac{1}{1+x^2}dx\]
\[\ x=e^u\]
\[\ dx=(du)e^u\]
\[\int\limits_{-\infty}^{0}\frac{e^u}{1+e^{2u}}du=\int\limits_{-\infty}^{0}\frac{1}{e^{-u}+e^{u}}du=\int\limits_{-\infty}^{0}\frac{1}{2cosh(u)}du\]
This question derives from an earlier one, and whilst I'd like to know what the aforestated integral is, are there are any erroneous statements above?

Answer 5

looks fine.

Answer 6

me too

Answer 7

Thanks- wolfram seemed to disagree with me earlier, but I may have mistyped.

Answer 8

So how would you actually do this, and does the cosh end make it any easier?

Answer 9

wolf has a funny way of "simplifying" things sometimes that confuses the hell out of me

Answer 10

Also- what cases in calculus CAN'T you just replace sinh with sine and at the end of the process, replace cosine with cosh? That is, do hyperbolic trig functions 'mirror' their hyperbolic peers in the same way as non-hyperbolic ones do?

Answer 11

yeah ... that was inverse of tangent. why do you want to change it into hyperbolic
http://www.wolframalpha.com/input/?i=integrate+1%2F%281%2Bx^2%29

Answer 12

I thought it could have been useful. How would you arrive at that using just integration laws?

Answer 13

that is one of standard trick ... d/dx arctan(x) = 1/(1+x^2) ... further more perhaps you could evaluate it using complex partial fraction.