Definite Integration Question 341
Question: $ \int_0^{1}{\frac{{{\tan }^{-1}}x}{1+x^{2}}}dx= $
[SCRA 1987; MNR 1990]
Options:
A) $ \frac{{{\pi }^{2}}}{8} $
B) $ \frac{{{\pi }^{2}}}{16} $
C) $ \frac{{{\pi }^{2}}}{4} $
D) $ \frac{{{\pi }^{2}}}{32} $
Show Answer
Answer:
Correct Answer: D
Solution:
Put $ t={{\tan }^{-1}}x\Rightarrow dt=\frac{1}{1+x^{2}}dx, $ then $ \int_0^{1}{\frac{{{\tan }^{-1}}x}{1+x^{2}}dx=\int_0^{\pi /4}{tdt=[ \frac{t^{2}}{2} ]_0^{\pi /4}=\frac{{{\pi }^{2}}}{32}}} $ .
BETA
