**Step 1: Use the trigonometric identity for sin(3x)** We know that $\sin 3x = 3\sin x - 4\sin^3 x$. So, the integral becomes: $\int \frac{\cos x}{3\sin x - 4\sin^3 x} dx$

**Step 2: Substitute u = sin(x)** Let $u = \sin x$. Then, $du = \cos x \, dx$. The integral transforms to: $\int \frac{1}{3u - 4u^3} du = \int \frac{1}{u(3 - 4u^2)} du$

**Step 3: Partial Fraction Decomposition** We need to decompose $\frac{1}{u(3 - 4u^2)}$ into partial fractions. Let $\frac{1}{u(3 - 4u^2)} = \frac{A}{u} + \frac{B}{ \sqrt{3} - 2u} + \frac{C}{\sqrt{3} + 2u}$ Multiplying both sides by $u(3 - 4u^2)$, we get: $1 = A(3 - 4u^2) + B(u(\sqrt{3} + 2u)) + C(u(\sqrt{3} - 2u))$ $1 = 3A - 4Au^2 + B\sqrt{3}u + 2Bu^2 + C\sqrt{3}u - 2Cu^2$ Comparing coefficients: - Constant term: $3A = 1 \Rightarrow A = \frac{1}{3}$ - $u$ term: $B\sqrt{3} + C\sqrt{3} = 0 \Rightarrow B = -C$ - $u^2$ term: $-4A + 2B - 2C = 0 \Rightarrow -4(\frac{1}{3}) + 2B - 2(-B) = 0 \Rightarrow -\frac{4}{3} + 4B = 0 \Rightarrow B = \frac{1}{3}$ So, $C = -\frac{1}{3}$. Thus, $\frac{1}{u(3 - 4u^2)} = \frac{1/3}{u} + \frac{1/3}{\sqrt{3} - 2u} - \frac{1/3}{\sqrt{3} + 2u}$

**Step 4: Integrate each term** $\int \frac{1}{u(3 - 4u^2)} du = \frac{1}{3} \int \frac{1}{u} du + \frac{1}{3} \int \frac{1}{\sqrt{3} - 2u} du - \frac{1}{3} \int \frac{1}{\sqrt{3} + 2u} du$ $= \frac{1}{3} \ln|u| + \frac{1}{3} \cdot \frac{-1}{2} \ln|\sqrt{3} - 2u| - \frac{1}{3} \cdot \frac{1}{2} \ln|\sqrt{3} + 2u| + C$ $= \frac{1}{3} \ln|u| - \frac{1}{6} \ln|\sqrt{3} - 2u| - \frac{1}{6} \ln|\sqrt{3} + 2u| + C$ $= \frac{1}{3} \ln|u| - \frac{1}{6} \ln|(\sqrt{3} - 2u)(\sqrt{3} + 2u)| + C$ $= \frac{1}{3} \ln|u| - \frac{1}{6} \ln|3 - 4u^2| + C$

**Step 5: Substitute back u = sin(x)** $\frac{1}{3} \ln|\sin x| - \frac{1}{6} \ln|3 - 4\sin^2 x| + C$ $= \frac{1}{6} [2\ln|\sin x| - \ln|3 - 4\sin^2 x|] + C$ $= \frac{1}{6} \ln\left|\frac{\sin^2 x}{3 - 4\sin^2 x}\right| + C$ Since $3 - 4\sin^2 x = 3 - 4(1 - \cos^2 x) = 4\cos^2 x - 1 = \cos 2x + 2\cos^2 x$, $\frac{1}{6} \ln\left|\frac{\sin^2 x}{3 - 4\sin^2 x}\right| + C = \frac{1}{6} \ln\left|\frac{\sin^2 x}{4\cos^2 x - 1}\right| + C$