A) \[{{\tan }^{-1}}\left( \frac{1-e}{1+e} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{e-1}{e+1} \right)\]
C) \[\frac{\pi }{4}\]
D) \[{{\tan }^{-1}}e+\frac{\pi }{4}\]
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 \[\int_{1}^{e}{\frac{dt}{1+{{t}^{2}}}=[{{\tan }^{-1}}t]_{1}^{e}}={{\tan }^{-1}}\left( \frac{e-1}{e+1} \right)\], \[\left[ \because {{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right) \right]\].
You need to login to perform this action.
You will be redirected in
3 sec
