Question

Q:-integrate this and solve
$∫ \sqrt{cotx} \times dx$

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Answer 1

Step-by-step explanation:

$\sf \int \sqrt{cotx} \:dx \\ \\ \sf Let\:cotx = {t}^{2} \\ \\ \sf Differentiating \:w.r.t \: x \\ \\ \sf \longrightarrow - {cosec}^{2}x = 2t \frac{dt}{dx} \\ \\ \sf \longrightarrow = - {cosec}^{2}x \:dx = 2t \: dt \\ \\ \sf \longrightarrow dx = \frac{-2t}{{cosec}^{2}x}\:dt$$\sf \longrightarrow dx = \frac {-2t}{1 +{cot}^{2}x}\:dt \\ \\ \sf \longrightarrow dx = \frac{-2t}{1+{{t}^{2}}^{2}}\:dt \\ \\ \sf \longrightarrow dx = \frac {-2t}{1+{t}^{4}}\:dt$$\sf \int \sqrt{cotx}\:dx = I \\ \\ \sf \longrightarrow I = - \int \sqrt{{t}^{2}} \:( \frac{2t}{1+{t}^{4}}) dt \\ \\ \sf \longrightarrow I = - \int (\frac {2{t}^{2}}{1+{t}^{4}}) dt \\ \\ \sf \longrightarrow I = - \int (\frac{2}{ {t}^{2} + \frac{1}{{t}^{2}}})dt$$\sf \longrightarrow I = - \int (\frac{ 1 + \frac{1}{{t}^{2}} + 1 - \frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}})dt \\ \\ \sf \longrightarrow - \int (\frac{1+\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}})dt - \int(\frac{1-\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}})dt$$\sf Now\: let \: {I}_{1} = \int (\frac{1+\frac{1}{{t}^{2}}}{{t}^{2}+\frac{1}{{t}^{2}}})dt \\ \\ \sf \longrightarrow {I}_{1} = \frac {1+\frac{1}{{t}^{2}}}{{(t+\frac{1}{t})}^{2}-2} \\ \\ \sf Let \: t - \frac{1}{t} = u \\ \\ \sf Differentiating\:w.r.t.\:t \\ \\ \sf \longrightarrow 1 + \frac{1}{{t}^{2}} = \frac{du}{dt}$$\sf \longrightarrow ( 1+ \frac{1}{{t}^{2}})dt = du \\ \\ \sf \longrightarrow {I}_{1} = \frac{du}{{u}^{2}+{\sqrt{2}}^{2}} \\ \\ \sf \longrightarrow {I}_{1} = \frac{1}{\sqrt{2}}{tan}^{-1}(\frac{u}{\sqrt{2}}) + {c}_{1} \\ \\ \sf \longrightarrow {I}_{1} = \frac{1}{\sqrt{2}}{tan}^{-1}(\frac{t-\frac{1}{t}}{\sqrt{2}}) + {c}_{1} \\ \\ \sf {I}_{1} = \frac{1}{\sqrt{2}}{tan}^{-1}(\frac{{t}^{2}-1}{\sqrt{2}t})+{c}_{1} \\ \\ \sf {I}_{2} = \int( \frac{1-\frac{1}{{t}^{2}}}{{t}^{2} + \frac{1}{{t}^{2}}})dt$$\sf \longrightarrow {I}_{2} \frac{ 1 - \frac{1}{{t}^{2}}}{{(t+\frac{1}{t})}^{2}-2} \\ \\ \sf Let\: t+\frac{1}{t} = v \\ \\ \sf Differenciating \: w.r.t. \:t \\ \\ \sf \longrightarrow 1 - \frac{1}{{t}^{2}} = \frac{dv}{dt} \\ \\ \sf \longrightarrow ( 1 - \frac{1}{{t}^{2}})dt = dv \\ \\ \sf {I}_{2} = \int \frac {dv}{{v}^{2}-{\sqrt{2}}^{2}} \\ \\ \sf \longrightarrow {I}_{2} = \frac{1}{2\sqrt{2}}log(\frac{v - \sqrt{2}}{v+\sqrt{2}}) + {c}_{2}$$\sf \longrightarrow {I}_{2} = \frac{1}{2\sqrt{2}}log(\frac{t + \frac{1}{t}- \sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}) + {c}_{2} \\ \\ \sf \longrightarrow {I}_{2} = \frac{1}{2\sqrt{2}}log(\frac{{t }^{2}+ 1- \sqrt{2}t}{{t}^{2}+1+\sqrt{2}t}) + {c}_{2} \\ \\ \sf I = -({I}_{1}+{I}_{2}) \\ \\ \sf \longrightarrow I = - ( \frac{1}{\sqrt{2}}{tan}^{-1}(\frac{{t}^{2}-1}{\sqrt{2}t}) + \frac{1}{2\sqrt{2}}log(\frac{{t }^{2}+ 1- \sqrt{2}t}{{t}^{2}+1+\sqrt{2}t}) ) + c$$\sf \longrightarrow I = - ( \frac{1}{\sqrt{2}}{tan}^{-1}(\frac{{cot }^{2}x-1}{\sqrt{2} cotx}) + \frac{1}{2\sqrt{2}}log(\frac{{cot }^{2}x+ 1- \sqrt{2}t}{{cot}^{2}x+1+\sqrt{2} cotx}) ) + c\\ \\ \sf Important\:formulas: \\ \\ \sf \star \star \:\:\int \frac{1}{{x}^{2}+{a}^{2}} = \frac{1}{a}{tan}^{-1}(\frac{x}{a}) \\ \\ \sf \star \star \:\: \int (\frac{1}{{x}^{2}-{a}^{2}})dx = \frac{1}{2a}log(\frac{x-a}{x+a})+c$

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Answer 2

$\green{\bold{\underline{ ☆ UPSC-ASPIRANT ☆} }}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-integrate this and solve

$∫ \sqrt{cotx} \times dx$

$\huge\tt\underline\blue{ANSWER }$

$⟹ - \frac{1}{ \sqrt{2} } {tan}^{ - 1} ( \frac{cotx - 1}{ \sqrt{2cotx} } ) - \frac{1}{2 \sqrt{2} } log( \frac{cotx + 1 - \sqrt{2cotx} }{cotx + 1 + \sqrt{2cotx} } ) + c$✓

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