Q.Choose the correct option: $\int \frac{1}{\cos x-\cos^2 x}dx =$
(A) $\log(\csc x-\cot x)+\tan\frac{x}{2}+c$ (B) $\sin 2x-\cos x+c$ (C) $\log(\sec x+\tan x)-\cot\frac{x}{2}+c$ (D) $\cos 2x-\sin x+c$

Write $\frac{1}{\cos x-\cos^2x}=\frac{1}{\cos x(1-\cos x)}$. Multiply numerator and denominator by $(1+\cos x)$: $\frac{1+\cos x}{\cos x(1-\cos^2x)}=\frac{1+\cos x}{\cos x\sin^2x}=\frac{1}{\cos x\sin^2x}+\csc^2x$. Now $\frac{1}{\cos x\sin^2x}=\frac{\sin^2x+\cos^2x}{\cos x\sin^2x}=\sec x+\frac{\cos x}{\sin^2x}$. So the full integrand is $\sec x+\frac{\cos x}{\sin^2x}+\csc^2x$. Integrating term by term: $\int\sec x\,dx=\log|\sec x+\tan x|$; $\int\frac{\cos x}{\sin^2x}dx=-\csc x$ (sub $u=\sin x$); $\int\csc^2x,dx=- …

Stopping at $-\csc x-\cot x$ without converting to the equivalent $-\cot\frac x2$ form, so the answer looks unlike a …