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Question Number 76298 by benjo last updated on 26/Dec/19

∫(√( tanx+cotxdx))

$$\int\sqrt{\:\mathrm{tanx}+\mathrm{cotxdx}} \\ $$

Commented by benjo last updated on 26/Dec/19

Commented by john santuy last updated on 26/Dec/19

using Weirstrras metode. let t=tan((x/2))

$${using}\:{Weirstrras}\:{metode}. \\ $$$${let}\:{t}={tan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$

Commented by benjo last updated on 26/Dec/19

please sir your write step by step

$$\mathrm{please}\:\mathrm{sir}\:\mathrm{your}\:\mathrm{write}\:\mathrm{step}\:\mathrm{by}\:\mathrm{step} \\ $$

Commented by mathmax by abdo last updated on 26/Dec/19

let I =∫ (√(tanx+(1/(tanx))))dx ⇒ I =∫(√(((sinx)/(cosx))+((cosx)/(sinx))))dx =∫(√(1/(cosxsinx)))dx =∫(1/(√((1/2)sin(2x))))dx =∫ ((√2)/(√(sin(2x))))dx vhangement (√(sin(2x)))=t give I =∫((√2)/t)((tdt)/(√(1−t^4 ))) because sin(2x)=t^2 ⇒2x =arcsin(t^2 ) ⇒2dx=((2tdt)/(√(1−t^4 ))) ⇒ I =∫ (((√2)dt)/(√(1−t^4 ))) and this integral is not resoluble by elementary functions..

$${let}\:{I}\:=\int\:\sqrt{{tanx}+\frac{\mathrm{1}}{{tanx}}}{dx}\:\Rightarrow\:{I}\:=\int\sqrt{\frac{{sinx}}{{cosx}}+\frac{{cosx}}{{sinx}}}{dx} \\ $$$$=\int\sqrt{\frac{\mathrm{1}}{{cosxsinx}}}{dx}\:=\int\frac{\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)}}{dx}\:=\int\:\frac{\sqrt{\mathrm{2}}}{\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx} \\ $$$${vhangement}\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}={t}\:{give}\:{I}\:=\int\frac{\sqrt{\mathrm{2}}}{{t}}\frac{{tdt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}\:{because} \\ $$$${sin}\left(\mathrm{2}{x}\right)={t}^{\mathrm{2}} \:\Rightarrow\mathrm{2}{x}\:={arcsin}\left({t}^{\mathrm{2}} \right)\:\Rightarrow\mathrm{2}{dx}=\frac{\mathrm{2}{tdt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }} \\ $$$$\Rightarrow\:{I}\:=\int\:\frac{\sqrt{\mathrm{2}}{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}\:{and}\:{this}\:{integral}\:{is}\:{not}\:{resoluble}\:{by}\:{elementary} \\ $$$${functions}.. \\ $$

Answered by john santuy last updated on 26/Dec/19

I=∫secx(√(cotx ))dx let cotx =u^2 →−csc^2 xdx=2u du dx = −((2u du)/(√(1+u^4 ))) I=∫((√(1+u^4 ))/u^2 )×u^2 ×((2u)/(√(1+u^4 ))) du = −2∫u du=−u^2 +C hence − cotx +C ■

$${I}=\int{secx}\sqrt{{cotx}\:}{dx} \\ $$$${let}\:{cotx}\:={u}^{\mathrm{2}} \rightarrow−{csc}^{\mathrm{2}} {xdx}=\mathrm{2}{u}\:{du} \\ $$$${dx}\:=\:−\frac{\mathrm{2}{u}\:{du}}{\sqrt{\mathrm{1}+{u}^{\mathrm{4}} }} \\ $$$${I}=\int\frac{\sqrt{\mathrm{1}+{u}^{\mathrm{4}} }}{{u}^{\mathrm{2}} }×{u}^{\mathrm{2}} \:×\frac{\mathrm{2}{u}}{\sqrt{\mathrm{1}+{u}^{\mathrm{4}} }}\:{du} \\ $$$$=\:−\mathrm{2}\int{u}\:{du}=−{u}^{\mathrm{2}} +{C} \\ $$$${hence}\:−\:{cotx}\:+{C}\:\blacksquare \\ $$

Commented by benjo last updated on 26/Dec/19

waw thanks sir

$$\mathrm{waw}\:\mathrm{thanks}\:\mathrm{sir} \\ $$

Commented by MJS last updated on 26/Dec/19

sorry but that′s wrong ∫(√(tan x +cot x))dx=∫(dx/(√(sin x cos x)))= =(√2)∫(dx/(√(sin 2x))) and this cannot be solved using elementary calculus

$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{that}'\mathrm{s}\:\mathrm{wrong} \\ $$$$\int\sqrt{\mathrm{tan}\:{x}\:+\mathrm{cot}\:{x}}{dx}=\int\frac{{dx}}{\sqrt{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}}= \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{dx}}{\sqrt{\mathrm{sin}\:\mathrm{2}{x}}} \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{elementary} \\ $$$$\mathrm{calculus} \\ $$

Commented by MJS last updated on 26/Dec/19

but (d/dx)[−cot x]=csc^2 x ≠ sec x (√(cot x))

$$\mathrm{but}\:\frac{{d}}{{dx}}\left[−\mathrm{cot}\:{x}\right]=\mathrm{csc}^{\mathrm{2}} \:{x}\:\neq\:\mathrm{sec}\:{x}\:\sqrt{\mathrm{cot}\:{x}} \\ $$

Commented by john santu last updated on 26/Dec/19

tanx +(1/(tanx))=((tan^2 x+1)/(tanx))=((sec^2 x)/(tanx)) (√((sec^2 x)/(tanx)))=secx(√(cotx )) sir?

$$\mathrm{tanx}\:+\frac{\mathrm{1}}{\mathrm{tanx}}=\frac{\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\mathrm{1}}{\mathrm{tanx}}=\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{tanx}} \\ $$$$\sqrt{\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\mathrm{tanx}}}=\mathrm{secx}\sqrt{\mathrm{cotx}\:}\:\mathrm{sir}? \\ $$

Commented by john santu last updated on 26/Dec/19

oo yes sir. i′m wrong in dx = −((2u)/(((√(1+u^2 )))^2 )) du

$${oo}\:{yes}\:{sir}.\:{i}'{m}\:{wrong}\:{in}\:{dx}\:=\:−\frac{\mathrm{2}{u}}{\left(\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:{du} \\ $$

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