Question

$\displaystyle \huge \bold \red{\int \sqrt{ \tan x} \: dx}$​

Answer 1

Answer:

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Answer 2

$\displaystyle \bold \red{\int \sqrt{ \tan x} \: dx}$

$\displaystyle \bold \red{ Let \: u , = \sqrt{ \tan x}}$

$\displaystyle \bold \red{ \: u {}^{2} = { \tan x}}$

$\displaystyle \bold \red{ \: u {}^{4} = { \tan^{2} x}}$

$\displaystyle \bold \red{ \: u {}^{4} = { \sec^{2} x - 1}}$

$\bold \red{u^4+1=sec {}^{2} x }$

$\bold \red{Since, 2udu=sec^2x \: dx}$

$\bold \red{dx = \frac{2udu}{ \sec^{2}x } }$

$\bold \red{dx = \frac{2udu}{ u^{4} + 1 } }$

${\displaystyle \bold \red{ \Rightarrow\int \sqrt{ \tan x} \: dx = \int u \frac{2u}{ {u}^{4} + 1 } du}}$

${\displaystyle \bold \red{ \Rightarrow \int \frac{2u {}^{2} }{ {u}^{4} + 1 } du}}$

Since, the power is very high, we need to reduce it.

${\displaystyle \bold \red{ \Rightarrow \int \frac{2u {}^{2} }{ {u}^{4} + 1 } du = \int \frac{ \frac{2 {u}^{2} }{ {u}^{2} } }{ \frac{ {u}^{4} }{ {u}^{2} } + \frac{1}{ {u}^{2} } }du = \int \frac{2}{ {u}^{2} + \frac{1}{ {u}^{2} } } du }}$

$\bold \red{Since, {u}^{2} + \frac{1}{ {u}^{2} } = ( {u})^{2} + ( \frac{1}{u} ) {}^{2} }$

${ \bold \red{= {u}^{2} + (2u)( \frac{1}{u} ) + ( \frac{1}{u} ) {}^{2} - 2(u)( \frac{1}{u} )}}$

$\bold \red{ = (u + \frac{1}{u} ) {}^{2} - 2}$

${ \displaystyle{ \bold \red{ \implies \int \frac{1 - \frac{1}{ {u}^{2} } }{(u + \frac{1}{u}) {}^{2} - 2} du + \int \frac{1 + \frac{1}{ {u}^{2} } }{(u - \frac{1}{u}) {}^{2} + 2 }du }}}$

${ \bold \red{Let, z=u+ \frac{1}{u} , dz = (1 - \frac{1}{ {u}^{2} })du }}$

$\bold \red{y=u− \frac{1}{u} ,dy=(1+ \frac{1}{ {u}^{2} } )du}$

${ \displaystyle{ \bold \red{ \implies \int \frac{1}{ {z}^{2} - 2} dz + \int \frac{1}{ {y}^{2} + 2} dy}}}$

${ \displaystyle{ \bold \red{ \implies \int \frac{1}{ {z}^{2} - ( \sqrt{2}) {}^{2} } dz + \int \frac{1}{ {y}^{2} + ( \sqrt{2} ) {}^{2} } dy}}}$

${ \bold \red {Since,  \displaystyle \int \frac{1}{ {a}^{2} - {b}^{2} } dx = - \frac{ \tanh {}^{ - 1} \frac{a}{b} }{b} + C}}$

${ \bold \red {and,  \displaystyle \int \frac{1}{ {a}^{2} + {b}^{2} } dx = \frac{ \tanh {}^{ - 1} \frac{a}{b} }{b} + C}}$

${\bold \red{⇒ \frac{ - 1}{ \sqrt{2} } \tanh {}^{ - 1} (\frac{z {}^{} }{ \sqrt{2} } )+ \frac{1}{ \sqrt{2} } \tan {}^{ - 1} ( \frac{y }{ \sqrt{2} } )}}$

${\bold \red{⇒ \frac{ - 1}{ \sqrt{2} } \tanh {}^{ - 1} \frac{(u + \frac{1}{u}) {}^{} }{ \sqrt{2} } + \frac{1}{ \sqrt{2} } \tan {}^{ - 1} \frac{(u - \frac{1}{u}) }{ \sqrt{2} } }}$

${\bold \red{⇒ \frac{ - 1}{ \sqrt{2} } \tanh {}^{ - 1}( \frac{ \sqrt{\tan x} + \sqrt{ \cot x} }{2}) + \frac{1}{ \sqrt{2} } \tan {}^{ - 1} ( \frac{ \sqrt{ \tan x} - \sqrt{ \cot x} }{ \sqrt{2} } ) +C }}$