Mathematics

Prove that $$\displaystyle\int^{\tan x}_{1/e}\dfrac{t}{1+t^2}dt+\displaystyle\int^{\cot x}_{1/e}\dfrac{1}{t(1+t^2)}dt=1$$.


SOLUTION
To prove $$\int _{ \cfrac { 1 }{ e }  }^{ \tan { x }  }{ \cfrac { t }{ 1+{ t }^{ 2 } } dt } +\int _{ \cfrac { 1 }{ e }  }^{ \cot { x }  }{ \cfrac { 1 }{ t(1+{ t }^{ 2 }) } dt } $$
Let $${ I }_{ 1 }=\int _{ \cfrac { 1 }{ e }  }^{ \tan { x }  }{ \cfrac { t }{ 1+{ t }^{ 2 } } dt } \\ { I }_{ 2 }=\int _{ \cfrac { 1 }{ e }  }^{ \cot { x }  }{ \cfrac { 1 }{ t(1+{ t }^{ 2 }) } dt } $$
and $$I={ I }_{ 1 }+{ I }_{ 2 }$$
also, $$\int _{ \cfrac { 1 }{ e }  }^{ \cot { x }  }{ \cfrac { 1 }{ t(1+{ t }^{ 2 }) } dt } $$
Let $$t=\cfrac { 1 }{ z } $$
$$dt=\cfrac { -1 }{ { z }^{ 2 } } dz$$
So, $${ I }_{ 1 }=\int _{ e }^{ \tan { x }  }{ \cfrac { -zdz }{ { z }^{ 2 }(1+\cfrac { 1 }{ { z }^{ 2 } } ) }  } =\int _{ \tan { x }  }^{ e }{ \cfrac { zdz }{ 1+{ z }^{ 2 } }  } $$
So, $${ I }_{ 2 }=\int _{ \tan { x }  }^{ e }{ \cfrac { tdt }{ 1+{ t }^{ 2 } }  } $$
So, $$I={ I }_{ 1 }+{ I }_{ 2 }$$
$$I=\int _{ \cfrac { 1 }{ e }  }^{ e }{ \cfrac { tdt }{ 1+{ t }^{ 2 } }  } \\ =\cfrac { 1 }{ 2 } \int _{ \cfrac { 1 }{ e }  }^{ e }{ \cfrac { 2t }{ 1+{ t }^{ 2 } } dt } \\ I=\cfrac { 1 }{ 2 } { \log { (1+{ t }^{ 2 }) }  }_{ \cfrac { 1 }{ e }  }^{ e }\\ I=\cfrac { 1 }{ 2 } \log { (1+{ e }^{ 2 }) } -\cfrac { 1 }{ 2 } \log { (1+\cfrac { 1 }{ { e }^{ 2 } } ) } \\ I=\cfrac { 1 }{ 2 } [\log { (\cfrac { 1+{ e }^{ 2 } }{ 1+\cfrac { 1 }{ { e }^{ 2 } }  } ) } ]\\ I=\cfrac { 1 }{ 2 } \log { { e }^{ 2 } } \\ I=\cfrac { 1 }{ 2 } \times 2\\ I=1$$
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Subjective Medium Published on 17th 09, 2020
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