1. **Stating the problem:** We want to find the half-range sine series for the piecewise function $$f(x) = \begin{cases} x, & 0 < x < 2 \\ 4 - x, & 2 < x < 4 \end{cases}$$ defined on the interval $(0,4)$. The half-range sine series represents $f(x)$ as a sum of sine terms on $(0,4)$. 2. **Formula for half-range sine series:** For $0 < x < L$, the half-range sine series is given by $$f(x) \sim \sum_{n=1}^\infty b_n \sin\left(\frac{n \pi x}{L}\right)$$ where $$b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n \pi x}{L}\right) dx$$ Here, $L=4$. 3. **Calculate coefficients $b_n$:** Split the integral according to the piecewise definition: $$b_n = \frac{2}{4} \left( \int_0^2 x \sin\left(\frac{n \pi x}{4}\right) dx + \int_2^4 (4 - x) \sin\left(\frac{n \pi x}{4}\right) dx \right)$$ 4. **Evaluate the first integral:** Let $I_1 = \int_0^2 x \sin\left(\frac{n \pi x}{4}\right) dx$. Use integration by parts: - Set $u = x$, $dv = \sin\left(\frac{n \pi x}{4}\right) dx$. - Then $du = dx$, and $v = -\frac{4}{n \pi} \cos\left(\frac{n \pi x}{4}\right)$. So, $$I_1 = -\frac{4x}{n \pi} \cos\left(\frac{n \pi x}{4}\right) \Big|_0^2 + \frac{4}{n \pi} \int_0^2 \cos\left(\frac{n \pi x}{4}\right) dx$$ The remaining integral: $$\int_0^2 \cos\left(\frac{n \pi x}{4}\right) dx = \frac{4}{n \pi} \sin\left(\frac{n \pi x}{4}\right) \Big|_0^2 = \frac{4}{n \pi} \sin\left(\frac{n \pi}{2}\right)$$ Therefore, $$I_1 = -\frac{8}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 5. **Evaluate the second integral:** Let $I_2 = \int_2^4 (4 - x) \sin\left(\frac{n \pi x}{4}\right) dx$. Rewrite as: $$I_2 = 4 \int_2^4 \sin\left(\frac{n \pi x}{4}\right) dx - \int_2^4 x \sin\left(\frac{n \pi x}{4}\right) dx$$ Calculate each separately: - First integral: $$\int_2^4 \sin\left(\frac{n \pi x}{4}\right) dx = -\frac{4}{n \pi} \cos\left(\frac{n \pi x}{4}\right) \Big|_2^4 = -\frac{4}{n \pi} \left( \cos(n \pi) - \cos\left(\frac{n \pi}{2}\right) \right)$$ - Second integral (similar to $I_1$ but limits 2 to 4): Use integration by parts again: $$\int_2^4 x \sin\left(\frac{n \pi x}{4}\right) dx = -\frac{4x}{n \pi} \cos\left(\frac{n \pi x}{4}\right) \Big|_2^4 + \frac{4}{n \pi} \int_2^4 \cos\left(\frac{n \pi x}{4}\right) dx$$ Calculate the remaining integral: $$\int_2^4 \cos\left(\frac{n \pi x}{4}\right) dx = \frac{4}{n \pi} \sin\left(\frac{n \pi x}{4}\right) \Big|_2^4 = \frac{4}{n \pi} \left( \sin(n \pi) - \sin\left(\frac{n \pi}{2}\right) \right) = -\frac{4}{n \pi} \sin\left(\frac{n \pi}{2}\right)$$ Putting it all together: $$\int_2^4 x \sin\left(\frac{n \pi x}{4}\right) dx = -\frac{16}{n \pi} \cos(n \pi) + \frac{8}{n \pi} \cos\left(\frac{n \pi}{2}\right) - \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 6. **Combine to find $I_2$:** $$I_2 = 4 \left(-\frac{4}{n \pi} \left( \cos(n \pi) - \cos\left(\frac{n \pi}{2}\right) \right) \right) - \left(-\frac{16}{n \pi} \cos(n \pi) + \frac{8}{n \pi} \cos\left(\frac{n \pi}{2}\right) - \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) \right)$$ Simplify: $$I_2 = -\frac{16}{n \pi} \cos(n \pi) + \frac{16}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{16}{n \pi} \cos(n \pi) - \frac{8}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ $$I_2 = \frac{8}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 7. **Calculate $b_n$:** $$b_n = \frac{1}{2} (I_1 + I_2) = \frac{1}{2} \left(-\frac{8}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) + \frac{8}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) \right)$$ Simplify: $$b_n = \frac{1}{2} \left( \frac{32}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) \right) = \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 8. **Final half-range sine series:** $$f(x) \sim \sum_{n=1}^\infty \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) \sin\left(\frac{n \pi x}{4}\right)$$ This series represents the given piecewise function on $(0,4)$ using sine terms only. **Answer:** $$b_n = \frac{16}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ and $$f(x) = \sum_{n=1}^\infty b_n \sin\left(\frac{n \pi x}{4}\right)$$