Definite Integration Question 28
Question: $ \int _{-1}^{1}{x{{\tan }^{-1}}xdx} $ equals
[RPET 1997]
Options:
A) $ ( \frac{\pi }{2}-1 ) $
B) $ ( \frac{\pi }{2}+1 ) $
C) $ (\pi -1) $
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int _{-1}^{1}{x{{\tan }^{-1}}xdx=2}\int_0^{1}{x{{\tan }^{-1}}xdx} $
$ \because x{{\tan }^{-1}}x $ is an even function $ I=[2\frac{x^{2}}{2}{{\tan }^{-1}}x]_0^{1}-2\int_0^{1}{\frac{1}{2}\frac{x^{2}}{1+x^{2}}dx} $
$ I=[x^{2}{{\tan }^{-1}}x]_0^{1}-\int_0^{1}{\frac{x^{2}+1-1}{1+x^{2}}dx} $
I = $ [x^{2}{{\tan }^{-1}}x]_0^{1}-[x]_0^{1}+[{{\tan }^{-1}}x]_0^{1} $
therefore $ I=\frac{\pi }{4}-1+\frac{\pi }{4}=\frac{\pi }{2}-1 $ .
BETA
