Question

Evaluate the integral? : `int_(-1)^1 coshx dx`

Answer 1

`int_(-1)^1 coshx dx = e-1/e " "(~~2.35040)`

Explanation:

`coshx` is an even function so its is symmetrical about `Oy`
graph{cosh x [-10, 10, -5, 5]}
So, `int_(-1)^1 coshx dx=2 int_0^1 coshx dx`

If we go back to the definition of `cosh x=1/2(e^x+e^-x)` we have:

`int_(-1)^1 coshx dx = 2 int_0^1 1/2(e^x+e^-x) dx`
`" " = int_0^1 (e^x+e^-x) dx`
`" " = [e^x-e^-x]_0^1`
`" " = (e^1-e^-1) - (e^0-e^0)`
`" " = e-1/e " "(~~2.35040)`

We could also use the fact that `int coshx=sinhx + C`