$\int\left\{\frac{(1-\cos x)}{[\cos x(1+\cos x)]}\right\} d x=$

$\log (\sec x+\tan x)-2 \tan \left(\frac{x}{2}\right)$

Here we have cos x but its d.c. i.e., -sin x is not present in the numerator and as such we cannot make the substitution of cos x = t. but we simply put cos x = t to split the integrand into partial fractions.

$\frac{1-\cos x}{\cos x(1+\cos x)}=\frac{1-t}{t(1+t)}$

$=\left(\frac{1}{t}-\frac{2}{1+t}\right)=\left(\frac{1}{\cos x}-\frac{2}{1+\cos x}\right)$

∴  $I=\int\left(\frac{1}{\cos x}-\frac{2}{1+\cos x}\right) d x=\int\left(\sec x-\sec ^2 \frac{x}{2}\right) dx$

$=\log (\sec x+\tan x)-2 \tan \left(\frac{x}{2}\right)$

Hence (1) is the correct answer.