Question Number 709 by malwaan last updated on 03/Mar/15
![find the integral in five different ways ∫[(x^2 −1)(x+1)]^(−(2/3)) dx](https://www.tinkutara.com/question/Q709.png)
$${find}\:{the}\:{integral}\:{in}\:{five}\:{different}\:{ways} \\ $$$$\int\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}+\mathrm{1}\right)\right]^{−\frac{\mathrm{2}}{\mathrm{3}}} {dx} \\ $$
Answered by prakash jain last updated on 03/Mar/15
![One solution (1/([(x^2 −1)(x+1)]^(2/3) ))=(1/([(x−1)(x+1)^2 ]^(2/3) )) =(1/((x−1)^(2/3) (x+1)^(1/3) (x+1))) =[((x+1)/(x−1))]^(2/3) ∙(1/((x+1)^2 )) ((x−1)/(x+1))=t⇒(2/((x+1)^2 ))dx=dt Integral becomrs ∫ t^(−2/3) ∙(dt/2)= (t^(1/3) /(1/3)) ∙(1/2)=(3/2)t^(1/3) =(3/2)[((x−1)/(x+1))]^(1/3) =(3/2)(((x−1)/(x+1)))^(1/3)](https://www.tinkutara.com/question/Q710.png)
$$\mathrm{One}\:\mathrm{solution} \\ $$$$\frac{\mathrm{1}}{\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}+\mathrm{1}\right)\right]^{\mathrm{2}/\mathrm{3}} }=\frac{\mathrm{1}}{\left[\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right]^{\mathrm{2}/\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \left({x}+\mathrm{1}\right)} \\ $$$$=\left[\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right]^{\mathrm{2}/\mathrm{3}} \centerdot\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}={t}\Rightarrow\frac{\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}={dt} \\ $$$$\mathrm{Integral}\:\mathrm{becomrs} \\ $$$$\int\:{t}^{−\mathrm{2}/\mathrm{3}} \centerdot\frac{{dt}}{\mathrm{2}}=\:\frac{{t}^{\mathrm{1}/\mathrm{3}} }{\mathrm{1}/\mathrm{3}}\:\centerdot\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{1}/\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{2}}\left[\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right]^{\mathrm{1}/\mathrm{3}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}} \\ $$
Commented by malwaan last updated on 03/Mar/15
$${thank}\:{you}\:{mr}\:{prakash}\: \\ $$$${and}\:{if}\:{you}\:{have}\:{another}\:{solution} \\ $$$${please}\:{send}\:{it}. \\ $$
Commented by prakash jain last updated on 03/Mar/15
![Note: result is valid only when the function [(x^2 −1)(x+1)]^(−2/3) takes real values. Some properties of exponents arr not valid for complex numbers.](https://www.tinkutara.com/question/Q714.png)
$$\mathrm{Note}:\:\mathrm{result}\:\mathrm{is}\:\mathrm{valid}\:\mathrm{only}\:\mathrm{when}\:\mathrm{the}\:\mathrm{function} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}+\mathrm{1}\right)\right]^{−\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{takes}\:\mathrm{real}\:\mathrm{values}.\:\mathrm{Some}\:\mathrm{properties}\:\mathrm{of}\:\mathrm{exponents} \\ $$$$\mathrm{arr}\:\mathrm{not}\:\mathrm{valid}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbers}. \\ $$
Answered by prakash jain last updated on 03/Mar/15
![x=tan^2 θ dx=2tan θ sec^2 θ dθ [(x^2 −1)(x+1)]^(−2/3) =[(tan^4 θ−1)(tan^2 θ+1)]^(−2/3) =[(tan^2 θ−1)sec^2 θsec^2 θ]^(−2/3) =[(((sin^2 θ−cos^2 θ)/(cos^2 θ)))sec^4 θ]^(−2/3) =[−cos 2θ∙sec^6 θ]^(−2/3) =(cos 2θ)^(−2/3) (sec θ)^(−4) Integral ∫(−cos 2θ)^(−2/3) (sec θ)^(−4) 2 tan θ sec^2 θ dθ ∫(−cos 2θ)^(−2/3) (2 ((tan θ)/(sec^2 θ))) dθ ∫(−cos 2θ)^(−2/3) (sin 2θ) dθ=(3/2)(−cos 2θ)^(1/3) =(3/2)(((x−1)/(x+1)))^(1/3)](https://www.tinkutara.com/question/Q715.png)
$${x}=\mathrm{tan}^{\mathrm{2}} \theta \\ $$$${dx}=\mathrm{2tan}\:\theta\:\mathrm{sec}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}+\mathrm{1}\right)\right]^{−\mathrm{2}/\mathrm{3}} \\ $$$$=\left[\left(\mathrm{tan}^{\mathrm{4}} \theta−\mathrm{1}\right)\left(\mathrm{tan}^{\mathrm{2}} \theta+\mathrm{1}\right)\right]^{−\mathrm{2}/\mathrm{3}} \\ $$$$=\left[\left(\mathrm{tan}^{\mathrm{2}} \theta−\mathrm{1}\right)\mathrm{sec}^{\mathrm{2}} \theta\mathrm{sec}^{\mathrm{2}} \theta\right]^{−\mathrm{2}/\mathrm{3}} \\ $$$$=\left[\left(\frac{\mathrm{sin}^{\mathrm{2}} \theta−\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{cos}^{\mathrm{2}} \theta}\right)\mathrm{sec}^{\mathrm{4}} \theta\right]^{−\mathrm{2}/\mathrm{3}} \\ $$$$=\left[−\mathrm{cos}\:\mathrm{2}\theta\centerdot\mathrm{sec}^{\mathrm{6}} \theta\right]^{−\mathrm{2}/\mathrm{3}} =\left(\mathrm{cos}\:\mathrm{2}\theta\right)^{−\mathrm{2}/\mathrm{3}} \left(\mathrm{sec}\:\theta\right)^{−\mathrm{4}} \\ $$$$\mathrm{Integral} \\ $$$$\int\left(−\mathrm{cos}\:\mathrm{2}\theta\right)^{−\mathrm{2}/\mathrm{3}} \left(\mathrm{sec}\:\theta\right)^{−\mathrm{4}} \:\mathrm{2}\:\mathrm{tan}\:\theta\:\mathrm{sec}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$\int\left(−\mathrm{cos}\:\mathrm{2}\theta\right)^{−\mathrm{2}/\mathrm{3}} \left(\mathrm{2}\:\frac{\mathrm{tan}\:\theta}{\mathrm{sec}^{\mathrm{2}} \theta}\right)\:{d}\theta \\ $$$$\int\left(−\mathrm{cos}\:\mathrm{2}\theta\right)^{−\mathrm{2}/\mathrm{3}} \left(\mathrm{sin}\:\mathrm{2}\theta\right)\:{d}\theta=\frac{\mathrm{3}}{\mathrm{2}}\left(−\mathrm{cos}\:\mathrm{2}\theta\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$
Commented by prakash jain last updated on 03/Mar/15
$$\mathrm{This}\:\mathrm{is}\:\mathrm{same}\:\mathrm{approach}\:\mathrm{as}\:\mathrm{previous}\:\mathrm{except} \\ $$$$\mathrm{a}\:\mathrm{different}\:\mathrm{substitution}\:\mathrm{has}\:\mathrm{been}\:\mathrm{used}. \\ $$
Commented by malwaan last updated on 03/Mar/15
$${thanks}\:{alot} \\ $$