Let's solve Problem 4 from your homework, which asks to find the exact length of the curve $y = e^x$ from $x = 0$ to $x = 1$, using the hint provided:

$\int \csc \theta \, d\theta = - \ln |\csc \theta + \cot \theta| + C$

Step 1: Formula for the length of a curve

The formula for the arc length of a curve $y = f(x)$ from $x = a$ to $x = b$ is:

$L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$

For the curve $y = e^x$, we have:

$\frac{dy}{dx} = \frac{d}{dx} \left( e^x \right) = e^x$

Step 2: Substituting into the formula

Now, we substitute $\frac{dy}{dx} = e^x$ into the arc length formula:

$L = \int_0^1 \sqrt{1 + (e^x)^2} \, dx = \int_0^1 \sqrt{1 + e^{2x}} \, dx$

Step 3: Use a trigonometric substitution

We can use the substitution $e^x = \tan \theta$, which implies that:

$dx = \frac{d\theta}{\sec^2 \theta}, \quad e^{2x} = \tan^2 \theta$

Thus, $1 + e^{2x} = 1 + \tan^2 \theta = \sec^2 \theta$, and the integral becomes:

$L = \int \sec \theta \, d\theta$

Step 4: Solve the integral

The integral of $\sec \theta$ is a known result:

$\int \sec \theta \, d\theta = \ln | \sec \theta + \tan \theta | + C$

Step 5: Back-substitute

Since $e^x = \tan \theta$, we have:

$\sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + e^{2x}}$

Thus, the arc length is:

$L = \left[ \ln \left| \sec \theta + \tan \theta \right| \right]_0^1$

Finally, evaluate the definite integral from $x = 0$ to $x = 1$, simplifying the result for the final answer.

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Would you like a detailed evaluation of the final answer, or any clarification on these steps?

Here are 5 questions that relate to this concept:

1. How does the arc length formula change for parametric equations?
2. Can you apply a similar method to find the length of $y = \ln(x)$?
3. What is the geometric interpretation of the result?
4. How can integration techniques simplify arc length calculations?
5. Why is trigonometric substitution effective in problems involving $e^x$?

Tip: Always check if trigonometric substitution or another method simplifies the integral before solving!