If $\int_0^1 \frac{c\sin^3 x}{4-\cos^2 x} dx = \pi\left[1-\frac{a\ln b}{c}\right]$ where $a$ and $b$ are prime and $c \in \mathbb{N}$, then $a+b+c = $ ____.
Step-by-Step Solution
Key Concept: Use substitution u = cos x to transform ∫₀¹ (sin³x)/(4-cos²x) dx into a rational function integral, then decompose using partial fractions on the resulting expression ∫ (1-u²)/((4-u²)(du)) to isolate logarithmic terms.
For $I = \int_0^\pi \frac{x \sin^3 x}{4 - \cos^2 x} dx$, use the property that $\int_0^\pi f(x) dx = \int_0^\pi f(\pi - x) dx$ to show $I = \pi \int_0^\pi \frac{\sin^3 x}{4 - \cos^2 x} dx / 2$. Then splitting the integral and using the symmetry of $\int_0^{\pi/2} \frac{\sin^3 x}{4 - \cos^2 x} dx$, substitute $\cos x = t$ to get $I = \pi \left[1 - \frac{3}{4}\ln 3\right]$. This gives $a = 1$, $b = 3$, $c = 4$, so $a + b + c = 10$.
Correct Answer: 10