Evaluate integral from -10 to 10 of ((2e^x)/(sinhx+coshx)dt)
Show steps please
Thanks!
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?
To evaluate the integral ∫((2e^x)/(sinhx+coshx)) dt from -10 to 10, we can follow these steps:
Step 1: Rewrite the integral using the exponential function.
The hyperbolic sine (sinhx) and hyperbolic cosine (coshx) functions can be written in terms of the exponential function (e^x) as:
sinhx = (e^x - e^(-x))/2
coshx = (e^x + e^(-x))/2
Step 2: Substitute the expressions for sinhx and coshx into the integral.
The integral becomes:
∫((2e^x)/((e^x - e^(-x))/2 + (e^x + e^(-x))/2)) dt
Step 3: Simplify the denominator.
Combine the fractions in the denominator:
∫((2e^x)/((2e^x)/2)) dt
Simplifying further:
∫dt
Step 4: Evaluate the integral.
Since dt is the differential of t, integrating dt with respect to t gives:
∫dt = t
Step 5: Plug in the limits of integration.
Now we need to evaluate t from -10 to 10:
[t] evaluated from -10 to 10 = 10 - (-10) = 20
Therefore, the value of the integral is 20.