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Question Number 171910 by Tawa11 last updated on 21/Jun/22
∫ (dx/(9 − 4x^2 )) using the trigonometric substitution.
$$\int\:\frac{\mathrm{dx}}{\mathrm{9}\:\:\:−\:\:\:\mathrm{4x}^{\mathrm{2}} } \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{substitution}. \\ $$
Answered by aleks041103 last updated on 21/Jun/22
∫(dx/(9−4x^2 ))=(1/9)∫(dx/(1−(((2x)/3))^2 ))=(1/9) (3/2)∫(du/(1−u^2 ))= =(1/6)∫(du/(1−u^2 )) (1/(1−u^2 ))=(1/((1−u)(1+u)))=(1/2)((1/(1−u))+(1/(1+u))) ∫(du/(1−u^2 ))=(1/2)ln∣1+u∣−(1/2)ln∣1−u∣ ⇒∫(dx/(9−4x^2 ))=(1/(12))ln∣((1+(2/3)x)/(1−(2/3)x))∣ ⇒∫(dx/(9−4x^2 ))=(1/(12))ln∣((3+2x)/(3−2x))∣
$$\int\frac{{dx}}{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{9}}\int\frac{{dx}}{\mathrm{1}−\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{9}}\:\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}+\frac{\mathrm{1}}{\mathrm{1}+{u}}\right) \\ $$$$\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}+{u}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}−{u}\mid \\ $$$$\Rightarrow\int\frac{{dx}}{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{12}}{ln}\mid\frac{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}{x}}\mid \\ $$$$\Rightarrow\int\frac{{dx}}{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{12}}{ln}\mid\frac{\mathrm{3}+\mathrm{2}{x}}{\mathrm{3}−\mathrm{2}{x}}\mid \\ $$
Commented by Tawa11 last updated on 21/Jun/22
I appreciate sir.
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 21/Jun/22
But sir, I need the trigonometric[substion. like. let x = 3 tanθ
$$\mathrm{But}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{need}\:\mathrm{the}\:\mathrm{trigonometric}\left[\mathrm{substion}.\right. \\ $$$$\mathrm{like}.\:\:\:\mathrm{let}\:\:\mathrm{x}\:\:=\:\:\mathrm{3}\:\mathrm{tan}\theta \\ $$
Commented by Tawa11 last updated on 21/Jun/22
Can the question be solved using trigonometric substitution?
$$\mathrm{Can}\:\mathrm{the}\:\mathrm{question}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{trigonometric}\:\mathrm{substitution}? \\ $$
Commented by mr W last updated on 21/Jun/22
certainly you can! that means you can always make easy things more complicated, when you like.
$${certainly}\:{you}\:{can}!\:{that}\:{means}\:{you} \\ $$$${can}\:{always}\:{make}\:{easy}\:{things}\:{more} \\ $$$${complicated},\:{when}\:{you}\:{like}. \\ $$
Commented by Tawa11 last updated on 21/Jun/22
Because the condition given is using trigonometric substitution sir.
$$\mathrm{Because}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{given}\:\mathrm{is}\:\mathrm{using}\:\mathrm{trigonometric}\:\mathrm{substitution}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 21/Jun/22
you can do it! i just said it is not a good idea. try with u=(3/2)tan θ, you will see.
$${you}\:{can}\:{do}\:{it}!\:{i}\:{just}\:{said}\:{it}\:{is}\:{not}\:{a} \\ $$$${good}\:{idea}.\:{try}\:{with}\:{u}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{tan}\:\theta,\:{you} \\ $$$${will}\:{see}. \\ $$
Commented by Tawa11 last updated on 21/Jun/22
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 22/Jun/22
∫ (dx/(9 − 4x^2 )) = ∫ (((3/2) sec^2 θ)/(9 − 9 tan^2 θ)) dθ = (3/(18)) ∫ ((sec^2 θ)/(1 − tan^2 θ)) dθ sec^2 θ = 1 + tan^2 θ From here, please what next ...
$$\int\:\frac{\mathrm{dx}}{\mathrm{9}\:\:−\:\:\mathrm{4x}^{\mathrm{2}} } \\ $$$$=\:\:\:\int\:\frac{\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{sec}^{\mathrm{2}} \:\theta}{\mathrm{9}\:\:−\:\:\mathrm{9}\:\mathrm{tan}^{\mathrm{2}} \theta}\:\mathrm{d}\theta \\ $$$$=\:\:\:\:\frac{\mathrm{3}}{\mathrm{18}}\:\int\:\frac{\mathrm{sec}^{\mathrm{2}} \theta}{\mathrm{1}\:\:−\:\:\mathrm{tan}^{\mathrm{2}} \theta}\:\:\mathrm{d}\theta \\ $$$$\mathrm{sec}^{\mathrm{2}} \theta\:\:\:=\:\:\:\mathrm{1}\:\:\:+\:\:\:\mathrm{tan}^{\mathrm{2}} \theta\:\:\:\:\:\:\: \\ $$$$\mathrm{From}\:\mathrm{here},\:\mathrm{please}\:\mathrm{what}\:\mathrm{next}\:… \\ $$
Answered by ajfour last updated on 22/Jun/22
let I=∫(dx/(9−4x^2 ))=4∫(dx/(36−16x^2 )) =4∫(dx/((6−4x)(6+4x))) =(4/(12))∫(((6−4x)+(6+4x))/((6−4x)(6+4x)))dx =(1/3)∫(dx/(6+4x))+(1/3)∫(dx/(6−4x)) =(1/(12))∫((4dx)/(6+4x))−(1/(12))∫ ((−4dx)/(6−4x)) =(1/(12))ln ∣6+4x∣−(1/(12))ln ∣6−4x∣+c =(1/(12))ln ∣((3+2x)/(3−2x))∣+c
$${let}\:\:\:{I}=\int\frac{{dx}}{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }=\mathrm{4}\int\frac{{dx}}{\mathrm{36}−\mathrm{16}{x}^{\mathrm{2}} } \\ $$$$\:\:\:=\mathrm{4}\int\frac{{dx}}{\left(\mathrm{6}−\mathrm{4}{x}\right)\left(\mathrm{6}+\mathrm{4}{x}\right)} \\ $$$$\:\:\:=\frac{\mathrm{4}}{\mathrm{12}}\int\frac{\left(\mathrm{6}−\mathrm{4}{x}\right)+\left(\mathrm{6}+\mathrm{4}{x}\right)}{\left(\mathrm{6}−\mathrm{4}{x}\right)\left(\mathrm{6}+\mathrm{4}{x}\right)}{dx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\mathrm{6}+\mathrm{4}{x}}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\mathrm{6}−\mathrm{4}{x}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{12}}\int\frac{\mathrm{4}{dx}}{\mathrm{6}+\mathrm{4}{x}}−\frac{\mathrm{1}}{\mathrm{12}}\int\:\frac{−\mathrm{4}{dx}}{\mathrm{6}−\mathrm{4}{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\:\mid\mathrm{6}+\mathrm{4}{x}\mid−\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\:\mid\mathrm{6}−\mathrm{4}{x}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\:\mid\frac{\mathrm{3}+\mathrm{2}{x}}{\mathrm{3}−\mathrm{2}{x}}\mid+{c} \\ $$
Commented by Tawa11 last updated on 22/Jun/22
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by thfchristopher last updated on 22/Jun/22
Let x=(3/2)sin θ dx=(3/2)cos θdθ sin θ=((2x)/3) , tan θ=((2x)/( (√(9−4x^2 )))) , sec θ=(3/( (√(9−4x^2 )))) ∴ ∫(dx/(9−4x^2 )) =(3/2)∫((cos θ)/(9cos^2 θ))dθ =(1/6)∫sec θdθ =(1/6)ln ∣tan θ+sec θ∣+C =(1/6)ln ∣((3+2x)/( (√(9−4x^2 ))))∣+C =(1/6)ln ∣(√((3+2x)/(3−2x)))∣+C =(1/(12))ln ∣((3+2x)/(3−2x))∣+C
$$\mathrm{Let}\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\theta \\ $$$${dx}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\:\theta{d}\theta \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{2}{x}}{\mathrm{3}}\:,\:\mathrm{tan}\:\theta=\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }}\:,\:\mathrm{sec}\:\theta=\frac{\mathrm{3}}{\:\sqrt{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }} \\ $$$$\therefore\:\int\frac{{dx}}{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{cos}\:\theta}{\mathrm{9cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{sec}\:\theta{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\mid\mathrm{tan}\:\theta+\mathrm{sec}\:\theta\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\mid\frac{\mathrm{3}+\mathrm{2}{x}}{\:\sqrt{\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} }}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\mid\sqrt{\frac{\mathrm{3}+\mathrm{2}{x}}{\mathrm{3}−\mathrm{2}{x}}}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\:\mid\frac{\mathrm{3}+\mathrm{2}{x}}{\mathrm{3}−\mathrm{2}{x}}\mid+{C} \\ $$
Commented by thfchristopher last updated on 22/Jun/22
Trigonometric substitution applied on integral format: (i)∫(dx/(a^2 +b^2 x^2 )) or ∫(dx/(b^2 x^2 +a^2 )) Let bx=atan θ (ii) ∫(dx/(a^2 −b^2 x^2 )) Let bx=asin θ (iii) ∫(dx/(b^2 x^2 −a^2 )) Let bx=asec θ
$$\mathrm{Trigonometric}\:\mathrm{substitution}\:\mathrm{applied}\:\mathrm{on}\: \\ $$$$\mathrm{integral}\:\mathrm{format}: \\ $$$$\left(\mathrm{i}\right)\int\frac{{dx}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} }\:\:\mathrm{or}\:\int\frac{{dx}}{{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:{bx}={a}\mathrm{tan}\:\theta \\ $$$$\left(\mathrm{ii}\right)\:\int\frac{{dx}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:{bx}={a}\mathrm{sin}\:\theta \\ $$$$\left(\mathrm{iii}\right)\:\int\frac{{dx}}{{b}^{\mathrm{2}} {x}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:{bx}={a}\mathrm{sec}\:\theta \\ $$
Commented by Tawa11 last updated on 22/Jun/22
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 22/Jun/22
I really appreciate.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Commented by mr W last updated on 22/Jun/22
a substitution doesn′t make the thing easier in any form, therefore unnecesary. e.g. ∫(dx/(a^2 −b^2 x^2 )) =(1/(2a))∫((1/(a−bx))+(1/(a+bx)))dx =(1/(2ab))[−ln (a−bx)+ln (a+bx)] =(1/(2ab))×ln ∣((a+bx)/(a−bx))∣+C ∫(dx/(b^2 x^2 −a^2 ))=−∫(dx/(a^2 −b^2 x^2 ))=−(1/(2ab))×ln ∣((a+bx)/(a−bx))∣+C
$${a}\:{substitution}\:{doesn}'{t}\:{make}\:{the}\:{thing} \\ $$$${easier}\:{in}\:{any}\:{form},\:{therefore}\:{unnecesary}. \\ $$$${e}.{g}. \\ $$$$\int\frac{{dx}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\int\left(\frac{\mathrm{1}}{{a}−{bx}}+\frac{\mathrm{1}}{{a}+{bx}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}\left[−\mathrm{ln}\:\left({a}−{bx}\right)+\mathrm{ln}\:\left({a}+{bx}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}×\mathrm{ln}\:\mid\frac{{a}+{bx}}{{a}−{bx}}\mid+{C} \\ $$$$\int\frac{{dx}}{{b}^{\mathrm{2}} {x}^{\mathrm{2}} −{a}^{\mathrm{2}} }=−\int\frac{{dx}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {x}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}{ab}}×\mathrm{ln}\:\mid\frac{{a}+{bx}}{{a}−{bx}}\mid+{C} \\ $$
Commented by Tawa11 last updated on 22/Jun/22
God bless you sir. I really appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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