Question

Question :-

$find \: \int [ \sqrt{cot \: x} \: + \sqrt{tan \: x} \: ] \: dx$
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Answer 1

$\bf We\:Have$

$\tt\to\int(\sqrt{Cotx} +\sqrt{Tanx} )dx$

$\bf We\:Know\:that$

$\tt\to Cotx=\dfrac{Cosx}{Sinx} \\\\\to\tt Tanx = \dfrac{Sinx}{Cosx}$

$\bf We\: Get$

$\tt\to\int\bigg(\dfrac{\sqrt{Cosx} }{\sqrt{Sinx} }+ \dfrac{\sqrt{Sinx} }{\sqrt{Cosx} } \bigg)dx$

$\tt\to \int \bigg(\dfrac{(\sqrt{Cosx})^2+(\sqrt{Sinx})^2 }{\sqrt{Sinx}\sqrt{Cosx}}\bigg)dx$

$\tt\to \int \bigg(\dfrac{Cosx+Sinx}{\sqrt{SinxCosx}}\bigg)dx$

$\bf Now\:Multiply\:and\:divide\:\sqrt{2}$

$\tt\to \int \sqrt{2} \bigg(\dfrac{Cosx+Sinx}{\sqrt{2SinxCosx}}\bigg)dx$

$\tt\to \int\dfrac {\sqrt{2}(Cosx+Sinx) }{\sqrt{Sin^2+Cos^2x-(Sin^2x+Cos^2x)+2SinxCosx} } dx$

$\tt\to\sqrt{2} \int\dfrac{(Cosx+Sinx)}{\sqrt{1 - (Sinx-Cosx)^2} } dx$

$\bf Now\:Using\:Substitution\:Method$

$\tt\to Sinx-Cosx=t$

$\tt\to \dfrac{d(Sinx-Cosx)}{dx} =\dfrac{dt}{dx}$

$\tt\to(Cosx+Sinx) =\dfrac{dt}{dx}$

$\tt\to(Sinx+Cosx)dx= dt$

$\bf Now\:Substitute \: Value$

$\tt\to\sqrt{2} \int\dfrac{dt}{\sqrt{1-t^2} }$

$\bf We\:Know\:That$

$\tt\to Sin^{-1}x=\dfrac{1}{\sqrt{1-x^2} }$

$\bf We\:Get$

$\tt\to\sqrt{2} (Sin^{-1}t) +c$

$\tt\to\sqrt{2} Sin^{-1}(Sinx-Cosx)+c$

$\bf Answer$

$\tt\to\sqrt{2} Sin^{-1}(Sinx-Cosx)+c$

Answer 2

Question :-

$find \: \int [ \sqrt{cot \: x} \: + \sqrt{tan \: x} \: ] \: dx$

solution:-

we have

${I}= \int [ \sqrt{cot \: x} \: + \sqrt{tan \: x} \: ] \: dx \: = \int \sqrt{tan} \: x(1 + cot \: x) \: dx \\ \\ put \: \: tan \: \: x \: = \: {t}^{2} , \: so \: \: that \: \: sec^{2} \: dx \: = \: 2t \: \: dt \\ \\ or \: \: \: \: \: dx \: \: = \frac{2t \: \: dt}{1 + {t}^{4} } \\ \\ Then \: \: \: \: {I} = \int \: t \: (1 + \frac{1}{t} ^{2} ) \: \frac{ \: \: \: 2t}{(1 +t^{4} ) } \: \: dt \\ \\ = 2$

$\int \frac{ ({t }^{2} + 1) }{ {t}^{4} + 1 } \: dt \: \: = 2 \int \frac{(1 + \frac{1}{ {t}^{2} })^{dt} }{( {t}^{2} + \frac{1}{ {t}^{2} }) } \\ \\ = 2 \int \frac{(1 + \frac{1}{ {t}^{2} } ) ^{dt} }{( t - \frac{1}{t} )^{2} \: \: + 2} \\ \\ put \: \: \: t - \frac{1}{t} = y \: ,so \: \: that \: \: (1 + \frac{1}{ {t}^{2} } ) \: dt \: = \: dy. \: \: Then$

$\\ \\ {I} \: = 2 \int \frac{dy}{ {y}^{2} + \sqrt{2} ^{2} } = \sqrt{2} \: tan^{ - 1} \: \frac{y}{ \sqrt{2} } + c = \sqrt{2} \: tan ^{ - 1} \: \frac{t - \frac{1}{t} }{ \sqrt{2} } + c= \sqrt{2} \: tan ^{ - 1} ( \frac{t ^{2} - 1 } { \sqrt{2 } \: t } ) + c= \sqrt{2} \: tan ^{ - 1} \: ( \frac{tan \: x \: - 1}{ \sqrt{2 \: tan \: x} } + c$