If $\int (\sin 3\theta + \sin \theta)^{\sin \theta} \cos d\theta = (A\sin^3 \theta + B\cos^2 \theta + C\sin \theta + D\cos \theta + E)e^{\sin \theta} + F$ then:
Step-by-Step Solution
Key Concept: Using sum-to-product formulas and integration by parts with exponential-trigonometric products reveals that the $\cos\theta$ term coefficient must equal zero by symmetry of the resulting expression.
First, simplify $\sin 3\theta + \sin \theta = 2\sin 2\theta \cos \theta$ using sum-to-product formula. The integral becomes $\int 2\sin 2\theta \cos \theta \cdot e^{\sin \theta} d\theta$. Using integration by parts with $u = 2\sin 2\theta$ and $dv = e^{\sin \theta}\cos \theta d\theta$, where $v = e^{\sin \theta}$. This gives $2\sin 2\theta e^{\sin \theta} - \int 4\cos 2\theta e^{\sin \theta}\cos\theta d\theta$. Since $\sin 2\theta = 2\sin\theta\cos\theta$, we get $4\sin\theta\cos\theta e^{\sin\theta}$. Continuing the process and comparing with the given form, we find $B = -12$, $D = 0$ (the $\cos\theta$ coefficient vanishes due to cancellation in repeated integration by parts).
Correct Answer: 2,3