Find the length of the curve over the given interval:

x=t+1
y=ln cos(t)
for t=0 ---> t=pi/4

To find the length of the curve, you can use the arc length formula. The formula for arc length of a curve parameterized by t is:

L = ∫√(dx/dt)^2 + (dy/dt)^2 dt

In this case, we have x = t + 1 and y = ln(cos(t)). To find dx/dt and dy/dt, we need to take the derivatives of x and y with respect to t.

dx/dt = d(t + 1)/dt = 1
dy/dt = d(ln(cos(t)))/dt

To evaluate dy/dt, we apply the chain rule. The derivative of ln(u) with respect to u is du/dt/u. In this case, u = cos(t), so the derivative of ln(cos(t)) with respect to t is:

dy/dt = -sin(t)/cos(t) = -tan(t)

Now we can substitute dx/dt = 1 and dy/dt = -tan(t) into the arc length formula:

L = ∫√(1^2 + (-tan(t))^2) dt
= ∫√(1 + tan^2(t)) dt
= ∫√sec^2(t) dt
= ∫sec(t) dt

To solve this integral, we can use the substitution method. Let u = sec(t), then du = sec(t)tan(t) dt. Rearranging, we have dt = du/(sec(t)tan(t)) = du/u.

Substituting u = sec(t) and dt = du/u into the integral, we have:

L = ∫sec(t) dt
= ∫(1/du) du
= ln|u| + C

Now, we need to find the limits of integration. The given interval is t = 0 to t = pi/4, so we substitute these values into the expression for u:

u(0) = sec(0) = 1
u(pi/4) = sec(pi/4) = √2

Substituting the limits of integration, we have:

L = ln|√2| - ln|1|
= ln(√2) - ln(1)
= ln(√2)

Therefore, the length of the curve over the given interval is ln(√2).