Definite Integration Question 40
Question: $ \int_0^{\pi /2}{\frac{x+\sin x}{1+\cos x}dx=} $
[MP PET 1989]
Options:
A) $ -\log 2 $
B) $ \log 2 $
C) $ \frac{\pi }{2} $
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Answer:
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} $ .
BETA
