To solve the integral \(\int \sin(\arctan(x)) \, dx\), we can approach the problem by using trigonometric identities and substitution. Here's the step-by-step solution:

### Step 1: Express \(\sin(\arctan(x))\) in a simpler form

Let \( \theta = \arctan(x) \). By definition of the arctangent, we have:
\[
\theta = \arctan(x) \implies \tan(\theta) = x
\]

Using the identity \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), we can represent this with a right triangle where the opposite side is \( x \) and the adjacent side is \( 1 \). The hypotenuse \( r \) can be calculated using the Pythagorean theorem:

\[
r = \sqrt{x^2 + 1}
\]

Now, \( \sin(\theta) \) is given by:

\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}}
\]

Therefore:

\[
\sin(\arctan(x)) = \frac{x}{\sqrt{x^2 + 1}}
\]

### Step 2: Substitute into the integral

Now we substitute this into the integral:

\[
\int \sin(\arctan(x)) \, dx = \int \frac{x}{\sqrt{x^2 + 1}} \, dx
\]

### Step 3: Solve the integral using substitution

Let \( u = x^2 + 1 \). Then \( du = 2x \, dx \), or \( \frac{du}{2} = x \, dx \).

The integral now becomes:

\[
\int \frac{x}{\sqrt{x^2 + 1}} \, dx = \frac{1}{2} \int \frac{du}{\sqrt{u}}
\]

### Step 4: Integrate

The integral \( \frac{1}{2} \int \frac{du}{\sqrt{u}} \) can be simplified as:

\[
\frac{1}{2} \int u^{-\frac{1}{2}} \, du = \frac{1}{2} \cdot 2u^{\frac{1}{2}} = \sqrt{u}
\]

Substituting back \( u = x^2 + 1 \):

\[
\sqrt{u} = \sqrt{x^2 + 1}
\]

Thus, the solution to the integral is:

\[
\int \sin(\arctan(x)) \, dx = \sqrt{x^2 + 1} + C
\]

where \( C \) is the constant of integration.

### Final Answer:
\[
\int \sin(\arctan(x)) \, dx = \sqrt{x^2 + 1} + C
\]

Would you like more details on any part of the solution, or do you have any questions?

### Relative Questions
1. What is the geometric interpretation of the trigonometric identity used here?
2. How would the approach change if the integral involved \(\cos(\arctan(x))\) instead?
3. Can we apply a similar method to evaluate \(\int \tan(\arctan(x)) \, dx\)?
4. What are some other common integrals involving inverse trigonometric functions?
5. How can we verify the result by differentiating the answer?

### Tip
When solving integrals involving inverse trigonometric functions, it's often helpful to first express the function in terms of a simpler trigonometric identity, using triangle relations.

Explore the step-by-step solution to the integral of sin(arctan(x)), including trigonometric identities and substitution methods. Suitable for advanced high school students and beyond.

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