$$
\begin{aligned}
& \text {Take } \mathrm{I}=\int \frac{\sin x \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} \cdot d x \\
& \mathrm{I}=\int \frac{\sin x \sec ^2 x-\tan x \cdot \sin x+\cos x}{1-1+2 \sin ^2 x} \cdot d x \\
& \mathrm{I}=\frac{1}{2}\left[\int \frac{\sec ^2 x}{\sin x} \cdot d x-\int \sec x d x+\int \frac{\cos x}{\sin ^2 x} \cdot d x\right] \\
& \text { Let } \sin x=t \\
& \cos x d x=d t \\
& \mathrm{I}=\frac{1}{2}\left[\int \frac{1+\tan ^2 x}{\sin x} \cdot d x-\log |\sec x+\tan x|+\int \frac{d t}{t^2}\right] \\
& \mathrm{I}=\frac{1}{2}\left[\int \operatorname{cosec} x d x+\int \sec x \tan x+d x-\log |\sec x+\tan x|-\frac{1}{t}\right]+c \\
& \mathrm{I}=\frac{1}{2}[\log |\operatorname{cosec} x-\cot x|+\sec x-\operatorname{cosec} x-\log |\sec x+\tan x|] \\
& I=\frac{1}{2}\left[\sec x-\operatorname{cosec} x+\log \left|\frac{2 \cos 2 x / 2}{2 \sin ^{x / 2} \cos ^{x / 2}}\right|-\log \left|\frac{\cos ^{x / 2}+\sin ^{x / 2}}{\cos ^{x / 2}-\sin ^{x / 2}}\right|\right] \\
& \mathrm{I}=\frac{1}{2}\left[\sec x-\operatorname{cosec} x+\log \left|\tan \frac{x}{2}\right|-\log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right|\right] \\
& \mathrm{I}=\frac{1}{2}\left[\sec x-\operatorname{cosec} x-\log \left|\frac{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}{\tan \frac{x}{2}}\right|\right]+c \text {. } \\
&
\end{aligned}
$$
So, option (c) is correct.