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 :
Let\[I=\int{\frac{x+\sin x}{1+\cos x}dx}\] \[=\int{\frac{x}{2{{\cos }^{2}}\frac{x}{2}}dx+\int{\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}}dx}}\] \[=\int{\frac{x}{2}{{\sec }^{2}}\frac{x}{2}dx+\int{\tan \frac{x}{2}dx}}\] \[=\frac{x}{2}\tan \frac{x}{2}-2\int{\tan \frac{x}{2}dx+\int{\tan \frac{x}{2}dx+}}\] \[=x\tan \frac{x}{2}+c\]
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