Mathematics

$\int_0^{\pi/2} \dfrac{tan^7 x}{cot^7 x + tan^7 x} dx$ is equal to

$\dfrac{\pi}{4}$

SOLUTION
$I=\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { \tan ^{ 7 }{ x } }{ \cot ^{ 7 }{ x } +\tan ^{ 7 }{ x } } } dx\longrightarrow 1$
$\int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f\left( a+b-x \right) } dx$
$\therefore I=\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { \tan ^{ 7 }{ \left( \cfrac { \pi }{ 2 } -x \right) } }{ \cot ^{ 7 }{ \left( \cfrac { \pi }{ 2 } -x \right) } +\tan ^{ 7 }{ \left( \cfrac { \pi }{ 2 } -x \right) } } }$
$I=\int _{ 0 }^{ { \pi }/{ 2 } }{ \cfrac { \cot ^{ 7 }{ x } }{ \tan ^{ 7 }{ x } +\cot ^{ 7 }{ x } } } dx\longrightarrow 2$
$\Longrightarrow 2I=\int _{ 0 }^{ { \pi }/{ 2 } }{ \left( \cfrac { \tan ^{ 7 }{ x } }{ \tan ^{ 7 }{ x } +\cot ^{ 7 }{ x } } +\cfrac { \cot ^{ 7 }{ x } }{ \tan ^{ 7 }{ x } +\cot ^{ 7 }{ x } } \right) }$
$\Longrightarrow 2I=\int _{ 0 }^{ { \pi }/{ 2 } }{ 1 } \cdot dx$
$\Longrightarrow 2I={ \left| x \right| }_{ 0 }^{ { \pi }/{ 2 } }$
$\Longrightarrow 2I=\cfrac { \pi }{ 2 }$
$I=\cfrac { \pi }{ 4 }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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