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tanx-cotxdx-




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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