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Calculus Level 2

The value of $I\quad =\quad \int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ { tan }^{ n+1 }x\quad dx\quad +\quad \cfrac { 1 }{ 2 } \int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ { tan }^{ n-1 }(\cfrac { x }{ 2 } } )\quad dx }$ is equal to

\cfrac { n+2 }{ 2n+1 }

\cfrac { 1 }{ n }

\cfrac { 2n-3 }{ 3n-2 }

\cfrac { 2n-1 }{ n }

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Suyash Mittal
Apr 25, 2014

$I\quad =\quad \int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ { tan }^{ n+1 }x\quad dx\quad +\quad \cfrac { 1 }{ 2 } \int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ { tan }^{ n-1 }(\cfrac { x }{ 2 } } )\quad dx }$ For second integral substitution $\cfrac { x }{ 2 } \quad =\quad y$ $I\quad =\quad \int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ ({ tan }^{ n+1 }x\quad +\quad { tan }^{ n-1 }x)dx }$ $\int _{ 0 }^{ \cfrac { \pi }{ 4 } }{ { tan }^{ n-1 }x\quad *\quad { sec }^{ 2 }x\quad dx }$ $[\cfrac { { tan }^{ n }x }{ n } \overset { \cfrac { \pi }{ 4 } }{ \underset { 0 }{ ] } } \quad =\quad \cfrac { 1 }{ n }$

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