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?
Evaluate integral from -10 to 10 of ((2e^x)/(sinhx+coshx)dt)
Show steps please
Thanks!
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