Definite-Integration Question 534
Question: $ \int_{,0}^{,\infty }{,\log ( x+\frac{1}{x} )\frac{dx}{1+x^{2}}} $ is equal to
[RPET 2000, 02]
Options:
A) $ \pi \log 2 $
B) $ -\pi \log 2 $
C) $ (\pi /2)\log 2 $
D) $ -(\pi /2)\log 2 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_0^{\infty }{\log ( x+\frac{1}{x} )}\frac{1}{1+x^{2}}dx $ Put $ x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta d\theta $
$ \Rightarrow I=\int_0^{\pi /2}{,\log (\tan \theta +\cot \theta })\frac{{{\sec }^{2}}\theta }{{{\sec }^{2}}\theta },d\theta $
Þ $ I=\int_0^{\pi /2}{,\log (\tan \theta +\cot \theta })d\theta $
$ \Rightarrow I=\int_0^{\pi /2}{\log \frac{(1+{{\tan }^{2}}\theta )}{\tan \theta },d\theta } $
Þ I $ =2\int_0^{\pi /2}{\log \sec \theta ,d\theta -\int_0^{\pi /2}{\log \tan \theta }},d\theta $
Þ I $ =2\int_0^{\pi /2}{\log \sec \theta d,\theta } $ ; $ { ,\because \int_0^{\pi /2}{\log \tan \theta =0} } $
$ \Rightarrow ,I=-2\int_0^{\pi /2}{,\log \cos \theta ,d\theta } $
Þ $ I=-2\times \frac{-\pi }{2}\log 2 $ , $ { \because \int_0^{\pi /2}{\log \cos \theta =-\frac{\pi }{2}\log 2} } $
Þ $ I=\pi \log 2 $ .
BETA
