G
Guest
Guest
The integral of 1/(e^x+e^-x) with respect to x(ie.dx)
thank you in advance[/tex]
thank you in advance[/tex]
Does it help you in knowing that \(\displaystyle \L\\\frac{1}{e^{x}+e^{-x}}=\frac{1}
{2}sech(x)\)
Probably not, huh?.
\(\displaystyle \L\\\int\frac{1}{e^{x}+e^{-x}}dx\)
Factor \(\displaystyle \L\\e^{x}\) out of the denominator:
\(\displaystyle \L\\\int\frac{1}{e^{x}(1+e^{-2x})}dx\)
Let \(\displaystyle \L\\u=-e^{-x};\;\ du=e^{-x}dx=\frac{1}{e^{x}}dx\)
\(\displaystyle \L\\\int\frac{1}{1+u^{2}}du\)
{2}sech(x)\)
Probably not, huh?.
\(\displaystyle \L\\\int\frac{1}{e^{x}+e^{-x}}dx\)
Factor \(\displaystyle \L\\e^{x}\) out of the denominator:
\(\displaystyle \L\\\int\frac{1}{e^{x}(1+e^{-2x})}dx\)
Let \(\displaystyle \L\\u=-e^{-x};\;\ du=e^{-x}dx=\frac{1}{e^{x}}dx\)
\(\displaystyle \L\\\int\frac{1}{1+u^{2}}du\)
skeeter
Elite Member
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- Dec 15, 2005
- Messages
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\(\displaystyle \L \frac{1}{e^x + e^{-x}} =\)
\(\displaystyle \L \frac{1}{\frac{e^{2x}}{e^x} + \frac{1}{e^x}} =\)
\(\displaystyle \L \frac{1}{\frac{e^{2x} + 1}{e^x}} =\)
\(\displaystyle \L \frac{e^x}{e^{2x} + 1}\)
\(\displaystyle \L \int \frac{e^x}{e^{2x} + 1} dx = arctan(e^x) + C\)
\(\displaystyle \L \frac{1}{\frac{e^{2x}}{e^x} + \frac{1}{e^x}} =\)
\(\displaystyle \L \frac{1}{\frac{e^{2x} + 1}{e^x}} =\)
\(\displaystyle \L \frac{e^x}{e^{2x} + 1}\)
\(\displaystyle \L \int \frac{e^x}{e^{2x} + 1} dx = arctan(e^x) + C\)