Definite Integration Question 377
Question: The value of $ \int _{1/e}^{\tan x}{\frac{tdt}{1+t^{2}}}+\int _{1/e}^{\cot x}{\frac{dt}{t(1+t^{2})}}= $
[IIT Screening]
Options:
A) $ -1 $
B) 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
On integrating both functions, we get $ =\frac{1}{2}| \log (1+t^{2}) | _{1/e}^{\tan x}+| { \log t-\frac{1}{2}\log (1+t^{2}) } | _{1/e}^{\cot x} $
$ =\frac{1}{2}[ \log {{\sec }^{2}}x-\log ( 1+\frac{1}{e^{2}} ) ]+\log \cot x-\log ( \frac{1}{e} ) $
$ -\frac{1}{2}{ \log (cose{c^{2}}x)-\log ( 1+\frac{1}{e^{2}} ) } $
$ =-\log ( \frac{1}{e} )=\log e=1 $ .
BETA
