Integral Calculus Question 226
Question: What is $ \int\limits_0^{1}{\frac{{{\tan }^{-1}}}{1+x^{2}}dx} $ equal to?
Options:
A) $ \frac{\pi }{4} $
B) $ \frac{\pi }{8} $
C) $ \frac{{{\pi }^{2}}}{8} $
D) $ \frac{{{\pi }^{2}}}{32} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let $ I=\int\limits_0^{1}{\frac{{{\tan }^{-1}}}{1+x^{2}}dx} $ Put $ {{\tan }^{-1}}x=t $ $ \frac{1}{1+x^{2}}dx=dt $ $ x=0,\Rightarrow t=0 $ $ x=1,\Rightarrow t=\pi /4 $
$ \therefore I=\int\limits_0^{\pi /4}{tdt}=. \frac{t^{2}}{2} |_0^{\pi /4}=\frac{{{\pi }^{2}}}{32} $
BETA
