Question

integral of
$\sqrt{tanx}$
​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\tt \: Evaluate \: \int \: \sqrt{tanx} \: dx$

$\large\underline{{ \bf \: Solution :- }}$

$\begin{gathered}\Large{\bold{{\underline{Formula\:Used - }}}} \end{gathered}$

$\boxed{ \tt \: \int\dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \bigg( \dfrac{x}{a} + c\bigg) + c}$

$\boxed{ \tt \: \int\dfrac{dx}{ {x}^{2} - {a}^{2} } = \dfrac{1}{</strong><strong>2</strong><strong>a} log\bigg(\dfrac{x - a}{x + a}\bigg) + c}$

Consider

$\rm :\longmapsto\: Let \: I \: = \int \: \sqrt{tanx}dx - - - (1)$

$\rm :\longmapsto\:\: \: \boxed{ \bf \:Put \: \sqrt{tanx} = y }$

$\rm :\implies\:tanx = {y}^{2}$

$\rm :\implies\: {sec}^{2} x = 2y\dfrac{dy}{dx}$

$\rm :\implies\: {sec}^{2} x \: dx = 2ydy$

$\rm :\implies\:dx = \dfrac{2ydy}{ {sec}^{2}x }$

$\rm :\implies\:dx \: = \: \dfrac{2y \: dy}{1 + {tan}^{2}x }$

$\rm :\implies\:dx \: = \: \dfrac{2y \: dy}{1 + {y}^{4} }$

So,

Equation (1) can be rewritten as

$\rm :\longmapsto\: I = \int \: y \times \dfrac{2y}{ {y}^{4} + 1 } dy$

$\rm :\longmapsto\: I = \int \: \dfrac{ {2y}^{2} }{ {y}^{4} + 1 } dy$

$\rm :\longmapsto\: I = \int \: \dfrac{ {y}^{2} + {y}^{2} }{ {y}^{4} + 1} dy$

$\rm :\longmapsto\: I = \int \: \dfrac{ {y}^{2} + {y}^{2} + 1 - 1}{ {y}^{4} + 1} dy$

$\rm :\longmapsto\: I = \int \: \dfrac{( {y}^{2} + 1)+ ({y}^{2} - 1) }{ {y}^{4} + 1} dy$

$\rm :\longmapsto\: I = \int\dfrac{ {y}^{2} + 1}{ {y}^{4} + 1} dy + \int\dfrac{ {y}^{2} - 1 }{ {y}^{4} + 1 } dy$

$\boxed{ \tt \: Divide \: numerator \: and \: denominator \: by \: {y}^{2} }$

$\rm :\longmapsto\: I = \int\dfrac{1 + \dfrac{1}{ {y}^{2} } }{ {y}^{2} + \dfrac{1}{ {y}^{2} } } dy + \int\dfrac{1 - \dfrac{1}{ {y}^{2} } }{ {y}^{2} + \dfrac{1}{ {y}^{2} } } dy$

$\rm :\longmapsto\: I = \int\dfrac{1 + \dfrac{1}{ {y}^{2} } }{ {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 + 2} dy + \int\dfrac{1 - \dfrac{1}{ {y}^{2} } }{ {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 - 2} dy$

$\rm I = \int\dfrac{1 + \dfrac{1}{ {y}^{2} } }{ { \bigg (y - \dfrac{1}{y} \bigg)}^{2} + 2 } dy + \int\dfrac{1 - \dfrac{1}{ {y}^{2} } }{ { \bigg (y + \dfrac{1}{y} \bigg)}^{2} - 2 } dy - - (2)$

Now,

$\rm :\longmapsto\:Put \: y - \dfrac{1}{y} = t$

$\rm :\implies\:(1 + \dfrac{1}{ {y}^{2} } )dy = dt$

And

$\rm :\longmapsto\:Put \: y + \dfrac{1}{y} = z$

$\rm :\implies\:(1 - \dfrac{1}{ {y}^{2} } )dy = dz$

So, equation (2) can be rewritten as

$\tt \: I = \dfrac{dt}{ {t}^{2} + 2 } + \dfrac{dz}{ {z}^{2} - 2}$

$\tt \: I = \dfrac{dt}{ {t}^{2} + {( \sqrt{2}) }^{2} } + \dfrac{dz}{ {z}^{2} - {( \sqrt{2}) }^{2} }$

$\bf\implies \: I = \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \dfrac{t}{ \sqrt{2} } + \dfrac{1}{2 \sqrt{2} } log \bigg(\dfrac{z - \sqrt{2} }{z + \sqrt{2} } \bigg) + c$

$\bf\implies \: I = \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \dfrac{y - \dfrac{1}{y}}{ \sqrt{2} } + \dfrac{1}{2 \sqrt{2} } log \bigg(\dfrac{y + \dfrac{1}{y} - \sqrt{2} }{y + \dfrac{1}{y} + \sqrt{2} } \bigg) + c$

$\tt \: I = \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \bigg( \dfrac{ {y}^{2} - 1 }{y \sqrt{2} } \bigg) + \dfrac{1}{2 \sqrt{2}} log(\bigg( \dfrac{ {y}^{2} +1 - y \sqrt{2} }{ {y}^{2} + 1 + y\sqrt{2} } \bigg)) + c$

$\tt \: I = \dfrac{1}{ \sqrt{2} } {tan}^{ - 1}\bigg(\dfrac{tanx - 1}{ \sqrt{2tanx} } \bigg) + \dfrac{1}{2 \sqrt{2} } log\bigg( \dfrac{ tanx + 1 - \sqrt{2tanx} }{tanx + 1 + \sqrt{2tanx} } \bigg) + c$