# calculus help

Evaluate integral from -10 to 10 of ((2e^x)/(sinhx+coshx)dt)

Show steps please
Thanks!

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1. Recall that
sinhx = (e^x-e^-x)/2
coshx = (e^x+e^-x)/2
so,
sinhx + coshx = e^x

2e^x/(sinhx+coshx) = 2e^x/e^x = 2

makes things kinda simple, eh?

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posted by Steve

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