Math, asked by Anonymous, 10 hours ago

   \large\displaystyle   \sf \red{\int_{0}^{ \frac{\pi}{2} }  \sqrt{ tanx} \:   tan( \frac{x}{2} )    \: \sec( \frac{x}{2} )  \: dx}

Answers

Answered by LaeeqAhmed
3

 \displaystyle \int_{0}^{ \frac{\pi}{2} } \sqrt{  \tan x} \:  \tan( \frac{x}{2} ) \: \sec( \frac{x}{2} ) \: dx

 \blue{ \boxed{  \tan x =  \frac{2 \tan( \frac{x}{2} )}{1 -  \tan ^{2}( \frac{x}{2} ) } }}

 \displaystyle \int_{0}^{ \frac{\pi}{2} } \sqrt{     \frac{2 \tan( \frac{x}{2} )}{1 -  \tan ^{2}( \frac{x}{2} ) } } \:  \tan( \frac{x}{2} ) \: \sec( \frac{x}{2} ) \: dx \:  \: ...(1)

 \sf \purple{put}

 { \tan(  \frac{x}{2} ) } =  {t}^{2}

 \implies  \frac{1}{2}  \sec ^{2}  (\frac{x}{2} ) \: dx = 2t \: dt

\implies    \: dx =  \frac{4t \: dt}{ \sec ^{2}  (\frac{x}{2} )} \: ....(2)

 \sf \purple{but}

 \tan( \frac{x}{2} ) =  {t}^{2}

 \blue{ \boxed{ \sec ^{2}x -  \tan  ^{2}  = 1 }}

 \implies \sec ^{2}  (\frac{x}{2} ) - ( {t}^{2} ) ^{2}  = 1

 \implies \sec ^{2}  (\frac{x}{2} ) - {t}^{4}   = 1

\therefore \sec ^{2}  (\frac{x}{2} ) = 1 + {t}^{4}    \:  \: ...(3)

 \sf  \purple{also}

 \sec   (\frac{x}{2} ) = \sqrt{ 1 + {t}^{4}   } \:  \: ...(4)

 \sf \purple{substituting \:  (3)\: in \: (2)}

\implies    \: dx =  \frac{4t \: dt}{  1 +  {t}^{4} } \:  \: ...(5)

 \sf \purple{limits}

 \sf upper \: limit =  \tan(\frac{\frac{\pi}{2}}{2})=\tan( \frac{\pi}{4} )  =  {t}^{2}

  \therefore\sf upper \: limit   =  {t} = 1

 \sf lower \: limit =  \tan( 0)  =  {t}^{2}

\sf \therefore lower \: limit  =  {t} = 0

 \sf \purple{substituting \:  (5) \: and \: (4)\: in \: (1)}

\displaystyle \int_{0}^{1 }  \sqrt{     \frac{2 ( {t}^{2} )}{1 -  ( {t}^{2} ) ^{2} } } \:  ( {t}^{2} ) \: ( \sqrt{1 +  {t}^{4}} )\: (\frac{4t \: dt}{  1 +  {t}^{4} })

 \implies\displaystyle \int_{0}^{1 }  \sqrt{     \frac{2  }{1 -   {t}^{4}   } } \: (t) ( {t}^{2} ) \: (\frac{4t \: dt}{   \sqrt{1 +  {t}^{4} }})

 \implies\displaystyle 4 \sqrt{2} \int_{0}^{1 }      \frac{ {t}^{4}  }{  \sqrt{(1 -   {t}^{4} ) ( 1+  {t}^{4} )} } \: \: dt

 \implies\displaystyle 4 \sqrt{2} \int_{0}^{1 }      \frac{ {t}^{4}  }{  \sqrt{1 -   ({t}^{4} {)}^{2}  } } \: \: dt

 \blue{ \boxed{ \displaystyle  \int  \frac{x}{ \sqrt{1 -  {x}^{2} }}   =  -  \sqrt{1 -  {x}^{2}  }   \: +  \:I_c }}

 \implies\displaystyle{ 4 \sqrt{2}      \large[  \small -  \sqrt{1 -   ({t}^{4} {)}^{2}  } \large  ]_{0}^{1 } }

 \implies\displaystyle{ 4 \sqrt{2}     \large[  \small -  \sqrt{1 -   {t}^{8}  }  \large ]_{0}^{1 } }

 \implies\displaystyle4 \sqrt{2}         \large[  \small  - \sqrt{1 -   {(1)}^{8}  }  - (  - \sqrt{1 -   {(0)}^{8}  }) \large ]

\implies\displaystyle4 \sqrt{2}         \large[   \small0   + 1 \large ]

\implies\displaystyle4 \sqrt{2}         \large[   \small   1 \large ]

\therefore\displaystyle4 \sqrt{2}         \large

 \orange{ \therefore\displaystyle \int_{0}^{ \frac{\pi}{2} } \sqrt{  \tan x} \:  \tan( \frac{x}{2} ) \: \sec( \frac{x}{2} ) \: dx = 4 \sqrt{2} }

HOPE IT HELPS!!

Answered by OoAryanKingoO78
10

Answer:

 \displaystyle \int_{0}^{ \frac{\pi}{2} } \sqrt{  \tan x} \:  \tan( \frac{x}{2} ) \: \sec( \frac{x}{2} ) \: dx

 \blue{ \boxed{  \tan x =  \frac{2 \tan( \frac{x}{2} )}{1 -  \tan ^{2}( \frac{x}{2} ) } }}

 \displaystyle \int_{0}^{ \frac{\pi}{2} } \sqrt{     \frac{2 \tan( \frac{x}{2} )}{1 -  \tan ^{2}( \frac{x}{2} ) } } \:  \tan( \frac{x}{2} ) \: \sec( \frac{x}{2} ) \: dx \:  \: ...(1)

 \sf \purple{put}

 { \tan(  \frac{x}{2} ) } =  {t}^{2}

 \implies  \frac{1}{2}  \sec ^{2}  (\frac{x}{2} ) \: dx = 2t \: dt

\implies    \: dx =  \frac{4t \: dt}{ \sec ^{2}  (\frac{x}{2} )} \: ....(2)

 \sf \purple{but}

 \tan( \frac{x}{2} ) =  {t}^{2}

 \blue{ \boxed{ \sec ^{2}x -  \tan  ^{2}  = 1 }}

 \implies \sec ^{2}  (\frac{x}{2} ) - ( {t}^{2} ) ^{2}  = 1

 \implies \sec ^{2}  (\frac{x}{2} ) - {t}^{4}   = 1

\therefore \sec ^{2}  (\frac{x}{2} ) = 1 + {t}^{4}    \:  \: ...(3)

 \sf  \purple{also}

 \sec   (\frac{x}{2} ) = \sqrt{ 1 + {t}^{4}   } \:  \: ...(4)

 \sf \purple{substituting \:  (3)\: in \: (2)}

\implies    \: dx =  \frac{4t \: dt}{  1 +  {t}^{4} } \:  \: ...(5)

 \sf \purple{limits}

 \sf upper \: limit =  \tan(\frac{\frac{\pi}{2}}{2})=\tan( \frac{\pi}{4} )  =  {t}^{2}

  \therefore\sf upper \: limit   =  {t} = 1

 \sf lower \: limit =  \tan( 0)  =  {t}^{2}

\sf \therefore lower \: limit  =  {t} = 0

 \sf \purple{substituting \:  (5) \: and \: (4)\: in \: (1)}

\displaystyle \int_{0}^{1 }  \sqrt{     \frac{2 ( {t}^{2} )}{1 -  ( {t}^{2} ) ^{2} } } \:  ( {t}^{2} ) \: ( \sqrt{1 +  {t}^{4}} )\: (\frac{4t \: dt}{  1 +  {t}^{4} })

 \implies\displaystyle \int_{0}^{1 }  \sqrt{     \frac{2  }{1 -   {t}^{4}   } } \: (t) ( {t}^{2} ) \: (\frac{4t \: dt}{   \sqrt{1 +  {t}^{4} }})

 \implies\displaystyle 4 \sqrt{2} \int_{0}^{1 }      \frac{ {t}^{4}  }{  \sqrt{(1 -   {t}^{4} ) ( 1+  {t}^{4} )} } \: \: dt

 \implies\displaystyle 4 \sqrt{2} \int_{0}^{1 }      \frac{ {t}^{4}  }{  \sqrt{1 -   ({t}^{4} {)}^{2}  } } \: \: dt

 \blue{ \boxed{ \displaystyle  \int  \frac{x}{ \sqrt{1 -  {x}^{2} }}   =  -  \sqrt{1 -  {x}^{2}  }   \: +  \:I_c }}

 \implies\displaystyle{ 4 \sqrt{2}      \large[  \small -  \sqrt{1 -   ({t}^{4} {)}^{2}  } \large  ]_{0}^{1 } }

 \implies\displaystyle{ 4 \sqrt{2}     \large[  \small -  \sqrt{1 -   {t}^{8}  }  \large ]_{0}^{1 } }

 \implies\displaystyle4 \sqrt{2}         \large[  \small  - \sqrt{1 -   {(1)}^{8}  }  - (  - \sqrt{1 -   {(0)}^{8}  }) \large ]

\implies\displaystyle4 \sqrt{2}         \large[   \small0   + 1 \large ]

\implies\displaystyle4 \sqrt{2}         \large[   \small   1 \large ]

\therefore\displaystyle4 \sqrt{2}         \large

 \purple{ \therefore\displaystyle \int_{0}^{ \frac{\pi}{2} } \sqrt{  \tan x} \:  \tan( \frac{x}{2} ) \: \sec( \frac{x}{2} ) \: dx = 4 \sqrt{2} }

\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}

HOPE IT HELPS!!:)

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