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