Mathematics

The value of $$\displaystyle \int_{0}^{\pi /2}x\left ( \sqrt{\tan x}+\sqrt{cotx} \right )\: dx$$ is?


ANSWER

$$\displaystyle \frac{\pi ^{2}}{2\sqrt{2}}$$


SOLUTION
Let $$I=\displaystyle \int _{ 0 }^{ \pi /2 } x\left( \sqrt { \tan  x } +\sqrt { cotx }  \right) \: dx$$
$$I=\displaystyle \int _{ 0 }^{ \pi /2 } \left( \cfrac { \pi  }{ 2 } -x \right) \left( \sqrt { \tan  \left( \cfrac { \pi  }{ 2 } -x \right)  } +\sqrt { cot\left( \cfrac { \pi  }{ 2 } -x \right)  }  \right) \: dx\\ =\cfrac { \pi  }{ 2 } \displaystyle \int _{ 0 }^{ \pi /2 } \left( \sqrt { \tan  x } +\sqrt { cotx }  \right) \: dx-\displaystyle \int _{ 0 }^{ \pi /2 } x\left( \sqrt { \tan  x } +\sqrt { cotx }  \right) \: dx\\ =\cfrac { \pi  }{ 2 } \displaystyle \int _{ 0 }^{ \pi /2 } \left( \sqrt { \tan  x } +\sqrt { cotx }  \right) \: dx-I\\ \Rightarrow 2I=\cfrac { \pi  }{ 2 } \displaystyle \int _{ 0 }^{ \pi /2 } \left( \sqrt { \tan  x } +\sqrt { cotx }  \right) \: dx\\ \Rightarrow I=\cfrac { \pi  }{ 4 } \displaystyle \int _{ 0 }^{ \pi /2 } \left( \sqrt { \tan  x } +\sqrt { cotx }  \right) \: dx$$
Now 
Let $$J=\displaystyle \int { \left( \sqrt { \tan  x } +\sqrt { cotx }  \right) \: dx } =\displaystyle \int { \cfrac { \tan { x } +1 }{ \sqrt { \tan { x }  }  } dx } $$
Put $$\tan { x } ={ t }^{ 2 }\Rightarrow \sec ^{ 2 }{ x } dx=2tdt$$
$$J=\displaystyle \int { \cfrac { { t }^{ 2 }+1 }{ \sqrt { { t }^{ 2 } }  } .\cfrac { 2t }{ { t }^{ 4 }+1 } dt } =2\displaystyle \int { \cfrac { { t }^{ 2 }+1 }{ { t }^{ 4 }+1 } dt } \\ =2\displaystyle \int { \cfrac { 1+\cfrac { 1 }{ { t }^{ 2 } }  }{ { t }^{ 2 }+\cfrac { 1 }{ { t }^{ 2 } } -2+2 } dt } =2\displaystyle \int { \cfrac { 1+\cfrac { 1 }{ { t }^{ 2 } }  }{ { \left( t-\cfrac { 1 }{ t }  \right)  }^{ 2 }+{ \left( \sqrt { 2 }  \right)  }^{ 2 } } dt } $$
Put $$t-\cfrac { 1 }{ t } =u\Rightarrow \left( 1+\cfrac { 1 }{ { t }^{ 2 } }  \right) dt=du$$
$$J=2\displaystyle \int { \cfrac { du }{ { u }^{ 2 }+{ \left( \sqrt { 2 }  \right)  }^{ 2 } }  } =\cfrac { 2 }{ \sqrt { 2 }  } \tan ^{ -1 }{ \left( \cfrac { u }{ \sqrt { 2 }  }  \right)  } +c\\ =\sqrt { 2 } \tan ^{ -1 }{ \left( \cfrac { \left( \sqrt { \tan  x } +\sqrt { cotx }  \right)  }{ \sqrt { 2 }  }  \right)  } +c$$
Hence $$I=\cfrac { { \pi  }^{ 2 } }{ 2\sqrt { 2 }  } $$
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Single Correct Hard Published on 17th 09, 2020
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