Definite Integration Question 403
Question: The value of $ \int_0^{1}{\frac{dx}{e^{x}+{e^{-x}}}} $ is
[SCRA 1980]
Options:
A) $ {{\tan }^{-1}}( \frac{1-e}{1+e} ) $
B) $ {{\tan }^{-1}}( \frac{e-1}{e+1} ) $
C) $ \frac{\pi }{4} $
D) $ {{\tan }^{-1}}e+\frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_0^{1}{\frac{dx}{e^{x}+{e^{-x}}}=\int_0^{1}{\frac{e^{x}}{1+e^{2x}}dx}} $
Now put $ e^{x}=t\Rightarrow e^{x}dx=dt $
Also as $ x=0 $ to 1, $ t=1 $ to e, then reduced form is $ \int_1^{e}{\frac{dt}{1+t^{2}}=[{{\tan }^{-1}}t]_1^{e}}={{\tan }^{-1}}( \frac{e-1}{e+1} ) $ , $ [ \because {{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}( \frac{x-y}{1+xy} ) ] $ .
BETA
