Menu Close

Evaluate-dx-x-4-9-x-2-

Tinku Tara 0 Comments
Pin




Question Number 179405 by Acem last updated on 29/Oct/22
Evaluate ∫ (dx/(x^4 (√(9−x^2 ))))
$$\:{Evaluate}\:\int\:\frac{{dx}}{{x}^{\mathrm{4}} \:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$
Answered by greougoury555 last updated on 29/Oct/22
M=∫ (dx/(x^3 x^2 (√((9/x^2 )−1)))) dx M= ∫ (x^(−3) /(x^2 (√(9x^(−2) −1)))) dx let j^2 = 9x^(−2) −1 ⇒x^(−2) = (1/9)(j^2 +1) 2j dj =−18x^(−3) dx M= ∫ (((−(1/9)j)dj)/((1/9)(j^2 +1).j))=−∫ (dj/(j^2 +1)) M=− arctan ((√((9/x^2 ) −1)) )+c M= − arctan ((x/( (√(9−x^2 )))) )+ c
$$\:{M}=\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:{x}^{\mathrm{2}} \:\sqrt{\frac{\mathrm{9}}{{x}^{\mathrm{2}} }−\mathrm{1}}}\:{dx} \\ $$$$\:{M}=\:\int\:\frac{{x}^{−\mathrm{3}} }{{x}^{\mathrm{2}} \:\sqrt{\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}}}\:{dx} \\ $$$$\:{let}\:{j}^{\mathrm{2}} \:=\:\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}^{−\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{9}}\left({j}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\mathrm{2}{j}\:{dj}\:=−\mathrm{18}{x}^{−\mathrm{3}} \:{dx} \\ $$$$\:{M}=\:\int\:\frac{\left(−\frac{\mathrm{1}}{\mathrm{9}}{j}\right){dj}}{\frac{\mathrm{1}}{\mathrm{9}}\left({j}^{\mathrm{2}} +\mathrm{1}\right).{j}}=−\int\:\frac{{dj}}{{j}^{\mathrm{2}} +\mathrm{1}} \\ $$$${M}=−\:\mathrm{arctan}\:\left(\sqrt{\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\:−\mathrm{1}}\:\right)+{c} \\ $$$$\:{M}=\:−\:\mathrm{arctan}\:\left(\frac{{x}}{\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}\:\right)+\:{c}\: \\ $$$$ \\ $$
Commented by Acem last updated on 29/Oct/22
M= ∫ (x^(−3) /(x^2 (√(9x^(−2) −1)))) dx M= ∫ (((−(1/9)j)dj)/((1/9)(j^2 +1).j))=−∫ (dj/(j^2 +1)) x^2 ≠ (1/9) (j^2 + 1)
$$\:{M}=\:\int\:\frac{{x}^{−\mathrm{3}} }{{x}^{\mathrm{2}} \:\sqrt{\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}}}\:{dx} \\ $$$$\:{M}=\:\int\:\frac{\left(−\frac{\mathrm{1}}{\mathrm{9}}{j}\right){dj}}{\frac{\mathrm{1}}{\mathrm{9}}\left({j}^{\mathrm{2}} +\mathrm{1}\right).{j}}=−\int\:\frac{{dj}}{{j}^{\mathrm{2}} +\mathrm{1}} \\ $$$${x}^{\mathrm{2}} \:\neq\:\frac{\mathrm{1}}{\mathrm{9}}\:\left({j}^{\mathrm{2}} \:+\:\mathrm{1}\right) \\ $$
Commented by greougoury555 last updated on 29/Oct/22
hmmm...x^(−2) ^� not x^2
$${hmmm}…{x}^{−\mathrm{2}} \:\:\:\bar {\:}\:{not}\:{x}^{\mathrm{2}} \\ $$
Commented by Acem last updated on 29/Oct/22
sure x^(−2) , and something will change
$${sure}\:{x}^{−\mathrm{2}} ,\:{and}\:{something}\:{will}\:{change}\: \\ $$
Commented by Acem last updated on 29/Oct/22
By the way, your method is very good!
$${By}\:{the}\:{way},\:{your}\:{method}\:{is}\:{very}\:{good}! \\ $$
Commented by Acem last updated on 29/Oct/22
When you supposed j^2 = 9x^(−2) −1 you got j dj= −9 x^(−3) dx, x^(−2) = (1/9) (j^2 +1) you had to write M= ∫((x^(−3) x^(−2) )/( (√(9x^(−2) −1)))) dx then M= ((−1)/(81))∫(j^2 +1) dj= ((−1)/(81)) ((1/3)j^3 + j)+c M= ((−1)/(81)) ( ((√((9−x^2 )^3 ))/(3x^3 )) + ((√(9−x^2 ))/x))+ c
$${When}\:{you}\:{supposed}\:{j}^{\mathrm{2}} =\:\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}\:{you}\:{got} \\ $$$$\:{j}\:{dj}=\:−\mathrm{9}\:{x}^{−\mathrm{3}} \:{dx},\:{x}^{−\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{9}}\:\left({j}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:{you}\:{had}\:{to}\:{write}\:{M}=\:\int\frac{{x}^{−\mathrm{3}} \:{x}^{−\mathrm{2}} }{\:\sqrt{\mathrm{9}{x}^{−\mathrm{2}} −\mathrm{1}}}\:{dx}\:{then} \\ $$$${M}=\:\frac{−\mathrm{1}}{\mathrm{81}}\int\left({j}^{\mathrm{2}} +\mathrm{1}\right)\:{dj}=\:\frac{−\mathrm{1}}{\mathrm{81}}\:\left(\frac{\mathrm{1}}{\mathrm{3}}{j}^{\mathrm{3}} +\:{j}\right)+{c} \\ $$$$ \\ $$$$\:{M}=\:\frac{−\mathrm{1}}{\mathrm{81}}\:\left(\:\frac{\sqrt{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{3}{x}^{\mathrm{3}} }\:+\:\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}}\right)+\:{c} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22
I=∫(dx/(x^4 (√(9−x^2 ))))=∫(dx/(x^5 (√((9/x^2 )−1)))), t=(1/x^2 ) ⇒dt=−(2/x^3 )dx =−(1/2)∫(t/( (√(9t−1))))dt=−(1/(18))∫((9t−1+1)/( (√(9t−1))))dt =−(1/(18))∫((√(9t−1))+(1/( (√(9t−1)))))dt =−(1/(162))((2/3)(9t−1)^(3/2) +2(√(9t−1)))+C =−(1/(81))(((√((9−x^2 )^3 ))/(3x^3 ))+((√(9−x^2 ))/x))+C
$${I}=\int\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}=\int\frac{{dx}}{{x}^{\mathrm{5}} \sqrt{\frac{\mathrm{9}}{{x}^{\mathrm{2}} }−\mathrm{1}}},\:{t}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow{dt}=−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}}{\:\sqrt{\mathrm{9}{t}−\mathrm{1}}}{dt}=−\frac{\mathrm{1}}{\mathrm{18}}\int\frac{\mathrm{9}{t}−\mathrm{1}+\mathrm{1}}{\:\sqrt{\mathrm{9}{t}−\mathrm{1}}}{dt} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{18}}\int\left(\sqrt{\mathrm{9}{t}−\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{9}{t}−\mathrm{1}}}\right){dt} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{162}}\left(\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{9}{t}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{2}\sqrt{\mathrm{9}{t}−\mathrm{1}}\right)+{C} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{81}}\left(\frac{\sqrt{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{3}{x}^{\mathrm{3}} }+\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}}\right)+{C} \\ $$
Commented by Acem last updated on 29/Oct/22
It′s very good substitution!! Thanks
$${It}'{s}\:{very}\:{good}\:{substitution}!!\:{Thanks} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22
I=∫(dx/(x^4 (√(9−x^2 )))), x=3sinθ =(1/(81))∫((cosθ)/(sin^4 θ∣cosθ∣))dθ=((sgn(cosθ))/(81))∫(dθ/(sin^4 θ)) =((sgn(cosθ))/(81))∫cosec^2 θ(cot^2 θ+1)dθ =−((sgn(cosθ))/(81))(((cot^3 θ)/3)+cotθ)+C
$${I}=\int\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{\mathrm{9}−{x}^{\mathrm{2}} }},\:{x}=\mathrm{3sin}\theta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{81}}\int\frac{\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{4}} \theta\mid\mathrm{cos}\theta\mid}{d}\theta=\frac{\mathrm{sgn}\left(\mathrm{cos}\theta\right)}{\mathrm{81}}\int\frac{{d}\theta}{\mathrm{sin}^{\mathrm{4}} \theta} \\ $$$$\:\:=\frac{\mathrm{sgn}\left(\mathrm{cos}\theta\right)}{\mathrm{81}}\int\mathrm{cosec}^{\mathrm{2}} \theta\left(\mathrm{cot}^{\mathrm{2}} \theta+\mathrm{1}\right){d}\theta \\ $$$$\:\:=−\frac{\mathrm{sgn}\left(\mathrm{cos}\theta\right)}{\mathrm{81}}\left(\frac{\mathrm{cot}^{\mathrm{3}} \theta}{\mathrm{3}}+\mathrm{cot}\theta\right)+{C} \\ $$
Commented by Acem last updated on 29/Oct/22
This is my method. By the way as long as the integration is undefined, we can deem that ∣cos θ∣= cos θ but if it was limit we have to compt the limits and see it domain if ((−π)/2)≤ θ≤ (π/2) then ∣cos θ∣= cos θ, otherwise −cos θ Thanks for you
$${This}\:{is}\:{my}\:{method}.\:{By}\:{the}\:{way}\:{as}\:{long}\:{as} \\ $$$$\:{the}\:{integration}\:{is}\:{undefined},\:{we}\:{can}\:{deem}\:{that} \\ $$$$\:\mid\mathrm{cos}\:\theta\mid=\:\mathrm{cos}\:\theta\:{but}\:{if}\:{it}\:{was}\:{limit}\:{we}\:{have}\:{to}\:{compt} \\ $$$$\:{the}\:{limits}\:{and}\:{see}\:{it}\:{domain} \\ $$$$\:{if}\:\frac{−\pi}{\mathrm{2}}\leqslant\:\theta\leqslant\:\frac{\pi}{\mathrm{2}}\:{then}\:\mid\mathrm{cos}\:\theta\mid=\:\mathrm{cos}\:\theta,\:{otherwise}\:−\mathrm{cos}\:\theta \\ $$$$\:{Thanks}\:{for}\:{you} \\ $$
Commented by ARUNG_Brandon_MBU last updated on 29/Oct/22
OK thanks
Answered by Acem last updated on 29/Oct/22
x= 3 sin θ ⇒ dx= 3 cos θ dθ (√(9−x^2 ))= 3 (√(cos^2 θ))= 3 ∣cos θ∣= 3 cos θ^( Integ. undefined) I= ∫ ((3 cos θ)/(81 sin^4 θ 3 cos θ)) dθ= (1/(81))∫cosec^2 θ (cot^2 θ+1) dθ = ((−1)/(81)) ((1/3) cot^3 θ + cot θ)+ c ; { (((cot θ)^′ = − cosec^2 θ)),((∫cosec^2 θdθ= −cot θ)) :} I= ((−1)/(81)) ( ((√((9−x)^3 ))/(3 x^3 )) + ((√(9−x^2 ))/x) ) + c Explan. sin θ= (x/3) ⇒ adjacent= (√(9−x^2 )) Thanks for all who solved
$$\:{x}=\:\mathrm{3}\:\mathrm{sin}\:\theta\:\Rightarrow\:{dx}=\:\mathrm{3}\:\mathrm{cos}\:\theta\:{d}\theta \\ $$$$\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }=\:\mathrm{3}\:\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta}=\:\mathrm{3}\:\mid\mathrm{cos}\:\theta\mid=\:\mathrm{3}\:\mathrm{cos}\:\theta^{\:{Integ}.\:{undefined}} \\ $$$${I}=\:\int\:\frac{\mathrm{3}\:\mathrm{cos}\:\theta}{\mathrm{81}\:\mathrm{sin}^{\mathrm{4}} \:\theta\:\mathrm{3}\:\mathrm{cos}\:\theta}\:{d}\theta=\:\frac{\mathrm{1}}{\mathrm{81}}\int\mathrm{cosec}^{\mathrm{2}} \:\theta\:\left(\mathrm{cot}^{\mathrm{2}} \:\theta+\mathrm{1}\right)\:{d}\theta \\ $$$$\:=\:\frac{−\mathrm{1}}{\mathrm{81}}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{cot}^{\mathrm{3}} \:\theta\:+\:\mathrm{cot}\:\theta\right)+\:{c}\:;\:\begin{cases}{\left(\mathrm{cot}\:\theta\right)^{'} =\:−\:\mathrm{cosec}^{\mathrm{2}} \:\theta}\\{\int\mathrm{cosec}^{\mathrm{2}} \:\theta{d}\theta=\:−\mathrm{cot}\:\theta}\end{cases} \\ $$$$\:{I}=\:\frac{−\mathrm{1}}{\mathrm{81}}\:\left(\:\frac{\sqrt{\left(\mathrm{9}−{x}\right)^{\mathrm{3}} }}{\mathrm{3}\:{x}^{\mathrm{3}} }\:+\:\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}}\:\right)\:+\:{c} \\ $$$$\:{Explan}.\:\mathrm{sin}\:\theta=\:\frac{{x}}{\mathrm{3}}\:\Rightarrow\:{adjacent}=\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$$${Thanks}\:{for}\:{all}\:{who}\:{solved} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *