1. **Stating the problem:** We want to find the integral $$\int \sin(px) \cos(qx) \, dx$$ where $p \neq q$ and $p \neq -q$. 2. **Formula and important rule:** Use the product-to-sum identity for sine and cosine: $$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$$ This helps convert the product into a sum of sines, which are easier to integrate. 3. **Apply the identity:** $$\sin(px) \cos(qx) = \frac{1}{2} [\sin((p+q)x) + \sin((p-q)x)]$$ 4. **Rewrite the integral:** $$\int \sin(px) \cos(qx) \, dx = \int \frac{1}{2} [\sin((p+q)x) + \sin((p-q)x)] \, dx = \frac{1}{2} \int \sin((p+q)x) \, dx + \frac{1}{2} \int \sin((p-q)x) \, dx$$ 5. **Integrate each term:** Recall that $$\int \sin(kx) \, dx = -\frac{\cos(kx)}{k} + C$$ So, $$\int \sin((p+q)x) \, dx = -\frac{\cos((p+q)x)}{p+q} + C$$ and $$\int \sin((p-q)x) \, dx = -\frac{\cos((p-q)x)}{p-q} + C$$ 6. **Combine results:** $$\int \sin(px) \cos(qx) \, dx = \frac{1}{2} \left(-\frac{\cos((p+q)x)}{p+q} - \frac{\cos((p-q)x)}{p-q}\right) + C$$ 7. **Final answer:** $$\boxed{\int \sin(px) \cos(qx) \, dx = \frac{1}{2} \left(-\frac{\cos((p+q)x)}{p+q} - \frac{\cos((p-q)x)}{p-q}\right) + C}$$