Definite Integration Question 143
Question: The value of $ \int _{0}^{1}{\frac{{{\tan }^{-1}}x}{1+x^{2}}dx} $ is
[RPET 2001]
Options:
A) $ \pi /4 $
B) $ {{\pi }^{2}}/32 $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_0^{1}{\frac{{{\tan }^{-1}}x}{1+x^{2}}dx} $ ; Put $ {{\tan }^{-1}}x=t $
therefore $ \frac{1}{1+x^{2}}dx=dt $
$ \therefore I=\int _{0}^{\pi /4}{tdt} $
$ =[ \frac{t^{2}}{2} ]_0^{\pi /4} $
$ =\frac{{{\pi }^{2}}}{32} $ .
BETA
