A) \[2\tan \frac{x}{2}-x+c\]
B) \[\frac{1}{2}\tan \frac{x}{2}-x+c\]
C) \[x-\frac{1}{2}\tan \frac{x}{2}+c\]
D) \[x-2\tan \frac{x}{2}+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{\cos x-1}{\cos x+1}\,dx=-\int_{{}}^{{}}{{{\tan }^{2}}\frac{x}{2}\,dx}}\] \[=-\int{\left( {{\sec }^{2}}\frac{x}{2}-1 \right)\,dx}=\int{\left( 1-{{\sec }^{2}}\frac{x}{2} \right)dx=x-2\tan \frac{x}{2}+c}\].
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