Step 1: Reduce the fraction:
Since $\sin4x = 4\sin x\cos x\cos2x$, the integrand becomes $\frac{1}{4\cos x\cos 2x}$.
Step 2: Change the variable:
Write $\frac{1}{\cos x\cos2x} = \frac{\cos x}{\cos^2x\cos2x}$. With $u=\sin x$ this becomes $\frac{du}{(1-u^2)(1-2u^2)}$.
Step 3: Split the fraction:
Use $\frac{1}{(1-u^2)(1-2u^2)} = \frac{2}{1-2u^2}-\frac{1}{1-u^2}$. You can check this by combining the right side: $\frac{2(1-u^2)-(1-2u^2)}{(1-u^2)(1-2u^2)} = \frac{1}{(1-u^2)(1-2u^2)}$.
Step 4: Standard forms:
Recall $\int\frac{du}{1-k^2u^2}=\frac{1}{2k}\ln\left|\frac{1+ku}{1-ku}\right|$. With $k=\sqrt2$ the first piece gives $\frac{2}{2\sqrt2}\ln\frac{1+\sqrt2u}{1-\sqrt2u}$. With $k=1$ the second piece gives $-\frac12\ln\frac{1+u}{1-u}$.
Step 5: Assemble:
Multiply by $\frac14$ to get $\frac{1}{4\sqrt2}\ln\frac{1+\sqrt2\sin x}{1-\sqrt2\sin x}-\frac18\ln\frac{1+\sin x}{1-\sin x}$. Since $\frac12\ln\frac{1+\sin x}{1-\sin x}=\ln|\sec x+\tan x|$, the last term is $-\frac14\ln|\sec x+\tan x|$. This is option (B).
Final Answer:
Option (B) is the result.
\[ \boxed{\text{Option (B)}} \]