Definite Integration Question 119
Question: The value of $ \int _{-1}^{3}{{{\tan }^{-1}}( \frac{x}{x^{2}+1} )+{{\tan }^{-1}}( \frac{x^{2}+1}{x} )dx} $ is
[Karnataka CET 2000]
Options:
A) $ 2\pi $
B) $ \pi $
C) $ \frac{21}{5}\pi $
D) $ \frac{\pi }{4} $
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Answer:
Correct Answer: A
Solution:
$ I=\int _{-1}^{3}{[ {{\tan }^{-1}}( \frac{x}{x^{2}+1} )+{{\cot }^{-1}}( \frac{x}{x^{2}+1} ) ]dx} $
$ =\int _{-1}^{3}{( \frac{\pi }{2} )dx=[ \frac{\pi x}{2} ] _{-1}^{3}=2\pi } $ , $ ( \because {{\tan }^{-1}}(x)+{{\cot }^{-1}}(x)=\frac{\pi }{2} ) $ .
BETA
