Definite Integration Question 27
Question: $ \int_0^{\pi /4}{{{\tan }^{2}}xdx=} $
[Roorkee 1983, Pb. CET 2000]
Options:
A) $ 1-\frac{\pi }{4} $
B) $ 1+\frac{\pi }{4} $
C) $ \frac{\pi }{4}-1 $
D) $ \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_0^{\pi /4}{{{\tan }^{2}}xdx=\int_0^{\pi /4}{({{\sec }^{2}}x-1)dx}} $
$ =\int_0^{\pi /4}{{{\sec }^{2}}xdx-\int_0^{\pi /4}{1dx}} $ = $ [\tan x]_0^{\pi /4}-[x]_0^{\pi /4}=1-\frac{\pi }{4} $ .
BETA
