$\sin x =\large\frac{2 \tan x/2}{1+ \tan ^2 x/2}$
$\cos x =\large\frac{1- \tan^2 x/2}{1+ \tan ^2 x/2}$
=> $-\int \large\frac{1+ \tan^2 x/2}{1- \tan ^2 x/2- 2 \tan (x/2)}$$.dx$
=> $-\int \large\frac{\sec^2 x/2}{( \tan x/2)^2+ 2 \tan (x/2)-1}$$.dx$
$\tan (x/2)=t$ differentiate with x
$\sec^2 (x/2)dx= 2 dt$
=> $-\int \large\frac{2 dt}{(t+1)^2 -(\sqrt 2)^2}$
=> $\int \large\frac{2 dt}{(\sqrt 2)^2-(t+1)^2}$
=> $\large\frac{2}{2 \times \sqrt 2 } \log |\large\frac{t+1+\sqrt 2}{\sqrt 2-(t+1)}|+c$
$\large\frac{1}{\sqrt 2}$$ \log _e \bigg|\large \frac{\sec^2 (x/2)+ (\sqrt 2 +1)}{(\sqrt 2 +1) \sec ^2 (x/2)}\bigg|+c $
Hence a is the correct answer.