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The given problem asks to find the Fourier series coefficient \( b_1 \) for the function \( f(x) = x \cos(x) \) over the interval \( (-\pi, \pi) \).

To solve this, we need to compute the Fourier sine coefficient \( b_1 \) using the following formula for a periodic function:

\[
b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx
\]

For the first sine coefficient (\( b_1 \)), we set \( n = 1 \), so we need to evaluate:

\[
b_1 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos(x) \sin(x) \, dx
\]

### Step 1: Simplify the integrand
We can use the trigonometric identity:

\[
\cos(x) \sin(x) = \frac{1}{2} \sin(2x)
\]

Thus, the integral becomes:

\[
b_1 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cdot \frac{1}{2} \sin(2x) \, dx
\]

\[
b_1 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x \sin(2x) \, dx
\]

### Step 2: Solve the integral
To compute this integral, we can apply integration by parts. Let \( u = x \) and \( dv = \sin(2x) dx \). The derivative of \( u \) and the integral of \( dv \) are:

\[
du = dx, \quad v = -\frac{1}{2} \cos(2x)
\]

Now apply integration by parts:

\[
\int x \sin(2x) \, dx = -\frac{x}{2} \cos(2x) + \frac{1}{2} \int \cos(2x) \, dx
\]

The integral of \( \cos(2x) \) is:

\[
\int \cos(2x) \, dx = \frac{1}{2} \sin(2x)
\]

So the result becomes:

\[
\int x \sin(2x) \, dx = -\frac{x}{2} \cos(2x) + \frac{1}{4} \sin(2x)
\]

### Step 3: Evaluate the definite integral
We now evaluate this expression from \( -\pi \) to \( \pi \):

\[
\left[ -\frac{x}{2} \cos(2x) + \frac{1}{4} \sin(2x) \right]_{-\pi}^{\pi}
\]

At both \( x = \pi \) and \( x = -\pi \), the terms involving \( \sin(2x) \) and \( \cos(2x) \) vanish because \( \sin(2\pi) = \sin(-2\pi) = 0 \) and \( \cos(2\pi) = \cos(-2\pi) = 1 \).

Thus, the entire expression evaluates to 0:

\[
b_1 = 0
\]

### Conclusion:
The correct value of \( b_1 \) is \( 0 \), so the correct answer is the last option:

\[
b_1 = 0
\]

Would you like more details on any part of this solution? Here are five related questions you might find interesting:

1. What are the general forms of Fourier series coefficients for sine and cosine terms?
2. How does the orthogonality of sine and cosine functions affect Fourier coefficients?
3. What is the significance of integration by parts in evaluating trigonometric integrals?
4. Can Fourier series represent discontinuous functions? How?
5. How are Fourier series applied in signal processing or physics?

**Tip:** In Fourier series, the sine and cosine coefficients can reveal symmetry properties of the function. If the function is odd, cosine terms (aₙ) will be zero, and if it's even, sine terms (bₙ) will be zero.

If f(x) = x cos(x) in (-π , π) then find the Fourier series coefficient b₁.

Learn how to find the Fourier series coefficient b₁ for the function f(x) = x cos(x) over the interval (-π, π). This problem involves key concepts like integration by parts, Fourier series, and trigonometric identities. Suitable for undergraduate-level students.

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