Question-211797

Question Number 211797 by Spillover last updated on 21/Sep/24

Answered by Ghisom last updated on 21/Sep/24

∫(((√(cot x))−(√(tan x)))/(1+3sin 2x))dx= [t=(√(tan x))] =−2∫((t^2 −1)/(t^4 +6t^2 +1))dt= [u=((2t)/(t^2 +1))] =∫(du/(u^2 +1))=arctan u =arctan ((2t)/(t^2 +1)) = =arctan ((2(√(tan x)))/(1+tan x)) = =arctan ((2(√(cos x sin x)))/(cos x +sin x)) +C

$$\int\frac{\sqrt{\mathrm{cot}\:{x}}−\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\mathrm{3sin}\:\mathrm{2}{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\right] \\ $$$$=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\mathrm{arctan}\:{u}\:=\mathrm{arctan}\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:= \\ $$$$=\mathrm{arctan}\:\frac{\mathrm{2}\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\mathrm{tan}\:{x}}\:= \\ $$$$=\mathrm{arctan}\:\frac{\mathrm{2}\sqrt{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}\:+{C} \\ $$

Commented by TonyCWX08 last updated on 22/Sep/24

$${I}\:{beg}\:{your}\:{pardon},\:{but}\:{I}\:{think}\:{this}\:{is}\:{wrong}. \\ $$

Commented by Ghisom last updated on 22/Sep/24

I′m convinced it′s right but feel free to prove it′s wrong

$$\mathrm{I}'\mathrm{m}\:\mathrm{convinced}\:\mathrm{it}'\mathrm{s}\:\mathrm{right} \\ $$$$\mathrm{but}\:\mathrm{feel}\:\mathrm{free}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$

Commented by TonyCWX08 last updated on 22/Sep/24

It should be 1−t^2 after the substitution

$${It}\:{should}\:{be}\:\mathrm{1}−{t}^{\mathrm{2}} \:{after}\:{the}\:{substitution} \\ $$

Commented by Frix last updated on 22/Sep/24

It′s correct: 2∫((1−t^2 )/(t^4 +6t^2 +1))dt=−2∫((t^2 −1)/(t^4 +6t^2 +1))dt

$$\mathrm{It}'\mathrm{s}\:\mathrm{correct}: \\ $$$$\mathrm{2}\int\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{dt}=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$

Answered by TonyCWX08 last updated on 22/Sep/24

I=∫(((√(cot(x)))−(√(tan(x))))/(1+3sin(2x))) dx =∫(1/( (√(tan(x)))))(((1−tan(x))/(1+6sin(x)cos(x))))dx =∫(1/( (√(tan(x)))))(((1−tan(x))/(sec^2 (x)+6tan(x))))(sec^2 (x))dx =∫(1/( (√(tan(x)))))(((1−tan(x))/(tan^2 (x)+6tan(x)+1)))(sec^2 (x))dx let t = (√(tan(x))) dt = ((sec^2 (x)dx)/(2(√(tan(x))))) ⇒2tdt=sec^2 (x)dx =∫(1/t)(((1−t^2 )/(t^4 +6t^2 +1)))(2tdt) =2∫(((1−t^2 )/(t^4 +6t^2 +1)))dt =2∫(((1−(1/t^2 ))/(t^2 +6+(1/t^2 ))))dt =2∫(((1−(1/t^2 ))/((t+(1/t))^2 +4)))dt Let v = t+(1/t) dv =( 1−(1/t^2 ) )dt =2∫(dv/(v^2 +4)) =2∫((dv/(v^2 +4))) =2(((tan^(−1) ((v/2)))/2))+c =tan^(−1) ((1/2)(t+(1/t)))+c =tan^(−1) ((1/2)((√(tan(x)))+(1/( (√(tan(x)))))))+c =tan^(−1) ((1/2)((√(tan(x)))+(√(cot(x)))))+c =tan^(−1) ((((√(tan(x)))+(√(cot(x))))/2))+C

$${I}=\int\frac{\sqrt{{cot}\left({x}\right)}−\sqrt{{tan}\left({x}\right)}}{\mathrm{1}+\mathrm{3}{sin}\left(\mathrm{2}{x}\right)}\:{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{{tan}\left({x}\right)}}\left(\frac{\mathrm{1}−{tan}\left({x}\right)}{\mathrm{1}+\mathrm{6}{sin}\left({x}\right){cos}\left({x}\right)}\right){dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{{tan}\left({x}\right)}}\left(\frac{\mathrm{1}−{tan}\left({x}\right)}{{sec}^{\mathrm{2}} \left({x}\right)+\mathrm{6}{tan}\left({x}\right)}\right)\left({sec}^{\mathrm{2}} \left({x}\right)\right){dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{{tan}\left({x}\right)}}\left(\frac{\mathrm{1}−{tan}\left({x}\right)}{{tan}^{\mathrm{2}} \left({x}\right)+\mathrm{6}{tan}\left({x}\right)+\mathrm{1}}\right)\left({sec}^{\mathrm{2}} \left({x}\right)\right){dx} \\ $$$${let}\:{t}\:=\:\sqrt{{tan}\left({x}\right)} \\ $$$$\:\:\:\:{dt}\:=\:\frac{{sec}^{\mathrm{2}} \left({x}\right){dx}}{\mathrm{2}\sqrt{{tan}\left({x}\right)}}\:\Rightarrow\mathrm{2}{tdt}={sec}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$=\int\frac{\mathrm{1}}{{t}}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}\right)\left(\mathrm{2}{tdt}\right) \\ $$$$=\mathrm{2}\int\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$$=\mathrm{2}\int\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\mathrm{6}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\right){dt} \\ $$$$=\mathrm{2}\int\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{4}}\right){dt} \\ $$$${Let}\:{v}\:=\:{t}+\frac{\mathrm{1}}{{t}} \\ $$$$\:\:\:\:\:\:{dv}\:=\left(\:\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\right){dt}\: \\ $$$$=\mathrm{2}\int\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{4}} \\ $$$$=\mathrm{2}\int\left(\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{4}}\right) \\ $$$$=\mathrm{2}\left(\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{v}}{\mathrm{2}}\right)}{\mathrm{2}}\right)+{c} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left({t}+\frac{\mathrm{1}}{{t}}\right)\right)+{c} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{{tan}\left({x}\right)}+\frac{\mathrm{1}}{\:\sqrt{{tan}\left({x}\right)}}\right)\right)+{c} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{{tan}\left({x}\right)}+\sqrt{{cot}\left({x}\right)}\right)\right)+{c} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{tan}\left({x}\right)}+\sqrt{{cot}\left({x}\right)}}{\mathrm{2}}\right)+{C} \\ $$

Commented by TonyCWX08 last updated on 22/Sep/24

$${Here}\:{is}\:{my}\:{solution}. \\ $$$${Inspired}\:{by}\:{Dr}.\:{Ghisom} \\ $$

Commented by Frix last updated on 22/Sep/24

2∫((1−t^2 )/(t^4 +6t^2 +1))dt=2∫(((1/t^2 )−1)/(t^2 +6+(1/t^2 )))dt=−^! 2∫((1−(1/t^2 ))/(t^2 +6+(1/t^2 )))dt

$$\mathrm{2}\int\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\mathrm{2}\int\frac{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{6}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt}=\overset{!} {−}\mathrm{2}\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\mathrm{6}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$

Commented by Spillover last updated on 23/Sep/24