1. **Problem Statement:** Evaluate the integral $$\int_{-\pi}^{\pi} \sin mx \sin nx \, dx$$ for the cases: (i) $m = n \neq 0$ (ii) $m = n = 0$ (iii) $m \neq n$ 2. **Formula and Important Rules:** The integral of the product of sine functions over $[-\pi, \pi]$ is related to the orthogonality of sine functions. The key formula is: $$\int_{-\pi}^{\pi} \sin mx \sin nx \, dx = \begin{cases} 0 & \text{if } m \neq n \\ \pi & \text{if } m = n \neq 0 \end{cases}$$ Also, note that $\sin 0x = 0$, so the case $m=n=0$ needs special attention. 3. **Step-by-step Evaluation:** **(i) Case $m = n \neq 0$:** $$\int_{-\pi}^{\pi} \sin^2 mx \, dx = \int_{-\pi}^{\pi} \frac{1 - \cos 2mx}{2} \, dx$$ Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$. Evaluate: $$= \frac{1}{2} \int_{-\pi}^{\pi} 1 \, dx - \frac{1}{2} \int_{-\pi}^{\pi} \cos 2mx \, dx$$ The first integral: $$\int_{-\pi}^{\pi} 1 \, dx = 2\pi$$ The second integral: $$\int_{-\pi}^{\pi} \cos 2mx \, dx = \left[ \frac{\sin 2mx}{2m} \right]_{-\pi}^{\pi} = 0$$ since $\sin$ is zero at multiples of $\pi$. Therefore: $$\int_{-\pi}^{\pi} \sin^2 mx \, dx = \frac{1}{2} (2\pi - 0) = \pi$$ **(ii) Case $m = n = 0$:** Here, $\sin 0x = 0$, so $$\int_{-\pi}^{\pi} \sin 0 \cdot \sin 0 \, dx = \int_{-\pi}^{\pi} 0 \, dx = 0$$ **(iii) Case $m \neq n$:** Use the product-to-sum identity: $$\sin mx \sin nx = \frac{1}{2} [\cos (m-n)x - \cos (m+n)x]$$ Then: $$\int_{-\pi}^{\pi} \sin mx \sin nx \, dx = \frac{1}{2} \int_{-\pi}^{\pi} \cos (m-n)x \, dx - \frac{1}{2} \int_{-\pi}^{\pi} \cos (m+n)x \, dx$$ Each integral evaluates to zero because: $$\int_{-\pi}^{\pi} \cos kx \, dx = \left[ \frac{\sin kx}{k} \right]_{-\pi}^{\pi} = 0$$ for any integer $k \neq 0$. Therefore: $$\int_{-\pi}^{\pi} \sin mx \sin nx \, dx = 0$$ 4. **Final Answers:** - For $m = n \neq 0$, the integral equals $\pi$. - For $m = n = 0$, the integral equals $0$. - For $m \neq n$, the integral equals $0$.