Mathematics

If $$\displaystyle \int_{0}^{1}\frac{dt}{e^{t}+e^{-t}}=\tan^{-1}p$$ then $$p$$, equals 


ANSWER

$$\displaystyle \frac{e-1}{e+1}$$


SOLUTION
$$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dt }{ e^{ t }+e^{ -t } } =\int _{ 0 }^{ 1 } \frac { e^{ t }dt }{ 1+e^{ 2t } } $$
Substitute $${ e }^{ t }=u\Rightarrow { e }^{ t }dt=du$$
$$\displaystyle \therefore I=\int _{ 1 }^{ e }{ \frac { 1 }{ 1+{ u }^{ 2 } } du } ={ \left[ \tan ^{ -1 }{ u }  \right]  }_{ 1 }^{ e }$$
$$\displaystyle =\tan ^{ -1 }{ e } -\tan ^{ -1 }{ 1 } =\tan ^{ -1 }{ \left( \frac { e-1 }{ 1+e }  \right)  } $$
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Single Correct Medium Published on 17th 09, 2020
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