Determine the distance traveled by a particle moving along the curve $$ y = \sin(x) $$ from $$ x = 0 $$ to $$ x = \pi $$.

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To determine the distance traveled by a particle moving along the curve $$ y = \sin(x) $$ from $$ x = 0 $$ to $$ x = \pi $$, we need to calculate the arc length of the curve over this interval.

### **Step 1: Arc Length Formula**

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

$$
S = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
$$

For $$ y = \sin(x) $$:

$$
\frac{dy}{dx} = \cos(x)
$$

So, the arc length becomes:

$$
S = \int_{0}^{\pi} \sqrt{1 + \cos^2(x)} \, dx
$$

### **Step 2: Simplifying the Integral**

The integral $$ \int \sqrt{1 + \cos^2(x)} \, dx $$ does not have a straightforward elementary antiderivative. However, it can be expressed in terms of **Elliptic Integrals**, which are special functions used to evaluate integrals of this form.

First, simplify the integrand using a trigonometric identity:

$$
1 + \cos^2(x) = \frac{3 + \cos(2x)}{2}
$$

So,

$$
\sqrt{1 + \cos^2(x)} = \sqrt{\frac{3 + \cos(2x)}{2}} = \frac{\sqrt{3 + \cos(2x)}}{\sqrt{2}}
$$

Thus, the arc length integral becomes:

$$
S = \frac{1}{\sqrt{2}} \int_{0}^{\pi} \sqrt{3 + \cos(2x)} \, dx
$$

Let’s make a substitution $$ u = 2x $$, which transforms the limits from $$ 0 $$ to $$ 2\pi $$:

$$
S = \frac{1}{2\sqrt{2}} \int_{0}^{2\pi} \sqrt{3 + \cos(u)} \, du
$$

### **Step 3: Expressing in Terms of Elliptic Integrals**

The integral $$ \int_{0}^{2\pi} \sqrt{3 + \cos(u)} \, du $$ is a standard form that can be expressed using the **Complete Elliptic Integral of the Second Kind**, denoted as $$ E(k) $$:

$$
\int_{0}^{2\pi} \sqrt{a + b \cos(u)} \, du = 4 \sqrt{a + b} \, E\left(k\right)
$$

where $$ k = \sqrt{\frac{2b}{a + b + \ldots}} $$.

Applying this to our integral with $$ a = 3 $$ and $$ b = 1 $$:

$$
\int_{0}^{2\pi} \sqrt{3 + \cos(u)} \, du = 8 \, E\left(\frac{1}{\sqrt{2}}\right)
$$

Therefore, the arc length $$ S $$ is:

$$
S = \frac{1}{2\sqrt{2}} \times 8 \, E\left(\frac{1}{\sqrt{2}}\right) = 2\sqrt{2} \, E\left(\frac{1}{\sqrt{2}}\right)
$$

### **Step 4: Numerical Approximation**

The complete elliptic integral $$ E\left(\frac{1}{\sqrt{2}}\right) $$ is approximately $$ 1.35064388 $$. Thus:

$$
S \approx 2 \times 1.41421356 \times 1.35064388 \approx 3.8212
$$

### **Final Answer**

The distance traveled is twice √2 times the complete elliptic integral E (1⁄√2). In symbols,

Distance = 2 √2 E(1/√2)

Numerically, this is approximately 3.82 units.
The distance traveled by the particle along the curve $$ y = \sin(x) $$ from $$ x = 0 $$ to $$ x = \pi $$ is approximately 3.82 units.

Determine the distance traveled by a particle moving along the curve from to .

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