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