Mathematics

# $\displaystyle \int_0^{\pi / 2} \dfrac{1}{1 + \sqrt [ 4]{tanx }}dx =$

$\pi / 4$

##### SOLUTION
$I=\int _{ 0 }^{ \pi }{ \cfrac { 1 }{ 1+{ \left( \tan { x } \right) }^{ 1/4 } } } dx=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { 1 }{ 1+{ \left( \tan { x } \right) }^{ 1/4 } } } dx\quad$
$2I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \left( \cfrac { 1 }{ 1+{ \left( \tan { x } \right) }^{ 1/4 } } +\cfrac { { \left( \tan { x } \right) }^{ 1/4 } }{ 1+{ \left( \tan { x } \right) }^{ 1/4 } } \right) } dx=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \left( \cfrac { 1+{ \left( \tan { x } \right) }^{ 1/4 } }{ 1+{ \left( \tan { x } \right) }^{ 1/4 } } \right) } dx$
$2I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ 1 } dx\Rightarrow 2I=\cfrac { \pi }{ 2 }$
$\Rightarrow I=\cfrac { \pi }{ 4 }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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