A) \[-\log 2\]
B) \[\log 2\]
C) \[\frac{\pi }{2}\]
D) 0
Correct Answer: C
Solution :
\[\int_{0}^{\pi /2}{\frac{x+\sin x}{1+\cos x}dx=\int_{0}^{\pi /2}{\frac{x+\sin x}{2{{\cos }^{2}}\frac{x}{2}}dx}}\] \[=\frac{1}{2}\int_{0}^{\pi /2}{x{{\sec }^{2}}\frac{x}{2}}dx+\int_{0}^{\pi /2}{\tan \frac{x}{2}dx}\]. \[=\left| \,x\tan \frac{x}{2}\, \right|_{0}^{\pi /2}=\frac{\pi }{2}\tan \frac{\pi }{4}=\frac{\pi }{2}\].
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