Integral-Calculus Question 516
Question: The value of $ \int_0^{\pi }{ln(1+\cos x)dx} $ is
Options:
A) $ \frac{\pi }{2}\log 2 $
B) $ \pi \log 2 $
C) $ -\pi \log 2 $
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I=\int_0^{\pi }{\log (1+\cos x)dx=\int_0^{\pi }{\log (2{{\cos }^{2}}\frac{x}{2})dx}} $ $ =\int_0^{\pi }{(\log 2+2\log \cos \frac{x}{2})dx} $ $ =\int_0^{\pi }{\log 2dx+2\int_0^{\pi }{\log \cos \frac{x}{2}dx}} $ $ =\pi \log 2+2\int_0^{\pi /2}{(2\log \cos t)dt} $ where $ \frac{x}{2}=t $ $ =\pi \log 2+4( -\frac{\pi }{2}\log 2 ) $ $ =\pi \log 2-2\pi \log 2=-\pi \log 2 $ $ [ \int_0^{\pi /2}{\log \sin \theta d\theta =\int_0^{\pi /2}{\log \cos \theta d\theta =-\frac{\pi }{2}\log 2}} ] $
BETA
