I'm asked to find the arc length of the curve y=3+(1/2)sinh(2x) from 0 to 1. I figured it out up to the integral of sqr(1+(cosh(2x))^2) from 0 to 1, but I'm not sure how to go on to solve it.
To find the arc length of the curve y = 3 + (1/2)sinh(2x) from 0 to 1, you have correctly set up the integral as follows:
∫[0 to 1] sqrt(1 + (cosh(2x))^2) dx
To solve this integral, we can use a trigonometric substitution. Let's substitute cosh(2x) = u:
cosh(2x) = u
Differentiating both sides with respect to x, we get:
2sinh(2x) dx = du
=> sinh(2x) dx = (1/2) du
Now, rewrite the integral in terms of u:
∫sqrt(1 + u^2) (1/2) du
To evaluate this integral, we can make a further substitution. Let:
u = tanh(t)
Differentiating both sides with respect to t, we get:
du = sech(t)^2 dt
Now rewrite the integral in terms of t:
∫sqrt(1 + tanh(t)^2) sech(t)^2 dt
Simplify the integrand:
∫sqrt(sech(t)^2) dt
= ∫sech(t) dt
The integral of sech(t) is a well-known integral and can be evaluated as:
∫sech(t) dt = ln|sech(t) + tanh(t)| + C
Finally, substitute back to get the solution in terms of x:
ln|sech(t) + tanh(t)| + C
= ln|sech(atanh(u)) + tanh(atanh(u))| + C
= ln|sqrt(1 + u^2) + u| + C
Now, evaluate the integral from 0 to 1:
ln|sqrt(1 + 1^2) + 1| - ln|sqrt(1 + 0^2) + 0|
= ln|sqrt(2) + 1| - ln|1|
= ln(sqrt(2) + 1)
Therefore, the arc length of the curve y = 3 + (1/2)sinh(2x) from 0 to 1 is ln(sqrt(2) + 1).