Sure, let's go through this step by step.
We wish to find the arc length of the curve y = tan(x) from x = 2π/3 to x = 4*3π/3. In calculus, the arc length of a curve defined by y = f(x) from a to b is given by the integral:
∫ from a to b (sqrt{1+[dy/dx]^2} dx), where dy/dx is the derivative of the function y = tan(x).
Here the function is y = tan(x).
Step 1: Finding derivative of y, dy/dx
The derivative of y = tan(x) is sec^2(x).
Step 2: Substituting the derivative into the integral
Now we substitute this into the integral. So the integral becomes: ∫ from a to b (sqrt{1 + sec^2(x)} dx)
However, Please note that squaring sec^2(x), remains sec^2(x) not sec^4(x). This is a common mistake that students make. The square of a trigonometric function is just the function again, not the function to the 4th power.
Here a = 2π/3 and b = 4*3π/3
So the arc length integral we are dealing with is:
∫ from 2π/3 to 3π/4 (sqrt{1 + sec^2(x)} dx).
Essentially we have to transform this expression to get rid of the square root.
The integral becomes ∫ from 2π/3 to 3π/4 (1+sec^2(x))dx, which is the formula for the arc length of the curve y = tan(x).
Therefore, the correct integral for the arc length of the curve y = tan(x) from x = 2π/3 to x = 4*3π/3 is ∫ from 2π/3 to 3π/4 (1+sec^2(x))dx.
Please be careful with your trigonometric functions. You should always check the derivatives of trigonometric functions carefully when dealing with integrals involving arc lengths or areas.