If $I_{m,n} = \int \cos^m x \sin nx dx$, then $7I_{4,3} - 4I_{3,2} =$$
Step-by-Step Solution
Key Concept: Apply integration by parts strategically to $I_{4,3} = \int \cos^4 x \sin 3x dx$ and $I_{3,2} = \int \cos^3 x \sin 2x dx$, then use trigonometric product-to-sum identities to express the integrands in terms of combinations of these integrals, ultimately creating a linear system that yields $-\cos^4 x \cos 3x + C$ when solving $7I_{4,3} - 4I_{3,2}$.
Integration by parts is applied to $I_{4,3} = \int \cos^3 x \sin 3x dx$. Using the identity $\sin x \cos 3x = -\sin 2x + \sin 3x \cos x$, the integral is reduced to a combination of $I_{4,3}$, $I_{3,2}$, and lower-order terms. Solving the resulting relation $\frac{7}{3}I_{4,3} - \frac{4}{3}I_{3,2} = \frac{\cos 3x \cos^3 x}{3} + C$ yields the final answer.
Correct Answer: 3