Definite Integration Question 72
Question: $ \int _{0}^{\pi /2}{\sin 2x\log \tan xdx} $ is equal to
[Kerala (Engg.) 2002; AI CBSE 1990; Karnataka CET 1996, 98]
Options:
A) $ \pi $
B) $ \pi /2 $
C) 0
D) $ 2\pi $
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_0^{\pi /2}{\sin 2x\log \tan xdx} $ , $ I=\int_0^{\pi /2}{\sin 2( \frac{\pi }{2}-x )\log \tan ( \frac{\pi }{2}-x )dx} $ , $ [\because \int_0^{a}{f(x)dx=\int_0^{a}{f(a-x)dx]}} $
$ =\int_0^{\pi /2}{\sin 2x\log \cot xdx} $
$ =-\int_0^{\pi /2}{\sin 2x\log \tan xdx} $
$ \therefore I=-I $
therefore 2I = 0
$ \Rightarrow I=0. $
BETA
