Question Number 98182 by abdomathmax last updated on 12/Jun/20
$$\mathrm{find}\:\int\:\mathrm{x}^{\mathrm{2}} \sqrt{\frac{\mathrm{2}−\mathrm{x}}{\mathrm{2}+\mathrm{x}}}\mathrm{dx} \\ $$
Answered by MJS last updated on 12/Jun/20
![∫x^2 (√((2−x)/(2+x)))dx= [t=(√((2+x)/(2−x))) → dx=(1/2)(√((2+x)(2−x)^3 ))dt] =32∫(((t^2 −1)^2 )/((t^2 +1)^4 ))dt= [Ostrogradski] =((8t(3t^4 +4t^2 +9))/(3(t^2 +1)^3 ))+8∫(dt/(t^2 +1))= =((8t(3t^4 +4t^2 +9))/(3(t^2 +1)^3 ))+8arctan t = =(1/3)(x^2 −3x+8)(√(4−x^2 ))+8arctan ((√(2+x))/( (√(2−x)))) = =(1/3)(x^2 −3x+8)(√(4−x^2 ))+8arcsin ((√(2+x))/2) +C](https://www.tinkutara.com/question/Q98212.png)
$$\int{x}^{\mathrm{2}} \sqrt{\frac{\mathrm{2}−{x}}{\mathrm{2}+{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left(\mathrm{2}+{x}\right)\left(\mathrm{2}−{x}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=\mathrm{32}\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{\mathrm{8}{t}\left(\mathrm{3}{t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{9}\right)}{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\mathrm{8}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{8}{t}\left(\mathrm{3}{t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{9}\right)}{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\mathrm{8arctan}\:{t}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{8}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+\mathrm{8arctan}\:\frac{\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{8}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }+\mathrm{8arcsin}\:\frac{\sqrt{\mathrm{2}+{x}}}{\mathrm{2}}\:+{C} \\ $$
Commented by mathmax by abdo last updated on 12/Jun/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by 1549442205 last updated on 12/Jun/20
![Putting x=2cosϕ(ϕ∈[0;π])⇒dx=−2sinϕdϕ F=∫4cos^2 ϕ.tan(ϕ/2)(−2sinϕ)dϕ.Put tan(ϕ/2)=u ⇒du=(1/2)(1+u^2 )dϕ,sinϕ=((2u)/(1+u^2 )),cos ϕ=((1−u^2 )/(1+u^2 )) F=−8∫(((1−u^2 )/(1+u^2 )))^2 .u.((2u)/(1+u^2 )).((2du)/(1+u^2 ))=−32∫((u^2 (1−u^2 )^2 )/((1+u^2 )^4 ))du =−32∫((u^6 −2u^4 +u^2 )/((1+u^2 )^4 ))du=−32∫(((u^2 +1)^3 −5(u^2 +1)^2 +8(u^2 +1)−4)/((u^2 +1)^4 ))du =−32∫(du/(u^2 +1))+160∫(du/((u^2 +1)^2 ))−96∫(du/((u^2 +1)^3 ))+128∫(du/((u^2 +1)^4 ))=A+B+C+D Note that we have the following formula: I_n =∫(dt/((t^2 +a^2 )^n ))=(1/(2a^2 (n−1))).(t/((t^2 +a^2 )^(n−1) ))+(1/a^2 ).((2n−3)/(2n−2)).I_(n−1) Hence,B=160I_2 =160{(1/2).(u/(u^2 +1))+(1/2)∫(du/(u^2 +1))}=((80u)/(u^2 +1))+80arctan(u) C=−96I_3 =−96{(1/4).(u/((u^2 +1)^2 ))+(3/4)[(u/(2(u^2 +1)))+(1/2)arctan(u)]} =((−24u)/((u^2 +1)^2 ))−((36u)/(u^2 +1))−36arctan(u) D=128I_4 =128{(1/6).(u/((u^2 +1)^3 ))+(5/6)I_3 }=128{(u/(6(u^2 +1)^3 ))+(5/6)[(u/(4(u^2 +1)^2 ))+(((3u)/(8(u^2 +1)))+(3/8)arctan(u))]} =((64u)/(3(u^2 +1)^3 ))+((80u)/(3(u^2 +1)^2 ))+((80u)/(3(u^2 +1)))+((120)/3)arctan(u) Thus,F=−32arctan(u)+((80u)/(u^2 +1))+80arctan(u)−((24u)/((u^2 +1)^2 )) −((36u)/(u^2 +1))−36arctan(u)+((64u)/(3(u^2 +1)^3 ))+((80u)/(3(u^2 +1)^2 ))+((80u)/(3(u^2 +1)))+((120)/3)arctan(u) =((156)/3)arctan(u)+((224u)/(3(u^2 +1)))+((152u)/(3(u^2 +1)^2 ))+((64u)/(3(u^2 +1)^3 )) with u=tan(ϕ/2) and x=2cosϕ](https://www.tinkutara.com/question/Q98231.png)
$$\mathrm{Putting}\:\mathrm{x}=\mathrm{2cos}\varphi\left(\varphi\in\left[\mathrm{0};\pi\right]\right)\Rightarrow\mathrm{dx}=−\mathrm{2sin}\varphi\mathrm{d}\varphi \\ $$$$\mathrm{F}=\int\mathrm{4cos}^{\mathrm{2}} \varphi.\mathrm{tan}\frac{\varphi}{\mathrm{2}}\left(−\mathrm{2sin}\varphi\right)\mathrm{d}\varphi.\mathrm{Put}\:\mathrm{tan}\frac{\varphi}{\mathrm{2}}=\mathrm{u} \\ $$$$\Rightarrow\mathrm{du}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\mathrm{d}\varphi,\mathrm{sin}\varphi=\frac{\mathrm{2u}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} },\mathrm{cos}\:\varphi=\frac{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{u}^{\mathrm{2}} } \\ $$$$\mathrm{F}=−\mathrm{8}\int\left(\frac{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\right)^{\mathrm{2}} .\mathrm{u}.\frac{\mathrm{2u}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }.\frac{\mathrm{2du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }=−\mathrm{32}\int\frac{\mathrm{u}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{4}} }\mathrm{du} \\ $$$$=−\mathrm{32}\int\frac{\mathrm{u}^{\mathrm{6}} −\mathrm{2u}^{\mathrm{4}} +\mathrm{u}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{4}} }\mathrm{du}=−\mathrm{32}\int\frac{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} −\mathrm{5}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{8}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{4}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }\mathrm{du} \\ $$$$=−\mathrm{32}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}+\mathrm{160}\int\frac{\mathrm{du}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{96}\int\frac{\mathrm{du}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\mathrm{128}\int\frac{\mathrm{du}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }=\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D} \\ $$$$\mathrm{Note}\:\mathrm{that}\:\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{following}\:\mathrm{formula}: \\ $$$$\:\mathrm{I}_{\mathrm{n}} =\int\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{n}} }=\frac{\mathrm{1}}{\mathrm{2a}^{\mathrm{2}} \left(\mathrm{n}−\mathrm{1}\right)}.\frac{\mathrm{t}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{n}−\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }.\frac{\mathrm{2n}−\mathrm{3}}{\mathrm{2n}−\mathrm{2}}.\mathrm{I}_{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{Hence},\mathrm{B}=\mathrm{160I}_{\mathrm{2}} =\mathrm{160}\left\{\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\right\}=\frac{\mathrm{80u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}+\mathrm{80arctan}\left(\mathrm{u}\right) \\ $$$$\mathrm{C}=−\mathrm{96I}_{\mathrm{3}} =−\mathrm{96}\left\{\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{u}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}}\left[\frac{\mathrm{u}}{\mathrm{2}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{u}\right)\right]\right\} \\ $$$$=\frac{−\mathrm{24u}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{36u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}−\mathrm{36arctan}\left(\mathrm{u}\right) \\ $$$$\mathrm{D}=\mathrm{128I}_{\mathrm{4}} =\mathrm{128}\left\{\frac{\mathrm{1}}{\mathrm{6}}.\frac{\mathrm{u}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{6}}\mathrm{I}_{\mathrm{3}} \right\}=\mathrm{128}\left\{\frac{\mathrm{u}}{\mathrm{6}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{6}}\left[\frac{\mathrm{u}}{\mathrm{4}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\left(\frac{\mathrm{3u}}{\mathrm{8}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{3}}{\mathrm{8}}\mathrm{arctan}\left(\mathrm{u}\right)\right)\right]\right\} \\ $$$$=\frac{\mathrm{64u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{80u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{80u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{120}}{\mathrm{3}}\mathrm{arctan}\left(\mathrm{u}\right) \\ $$$$\mathrm{Thus},\mathrm{F}=−\mathrm{32arctan}\left(\mathrm{u}\right)+\frac{\mathrm{80u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}+\mathrm{80arctan}\left(\mathrm{u}\right)−\frac{\mathrm{24u}}{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{36u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}−\mathrm{36arctan}\left(\mathrm{u}\right)+\frac{\mathrm{64u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{80u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{80u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{120}}{\mathrm{3}}\mathrm{arctan}\left(\mathrm{u}\right) \\ $$$$=\frac{\mathrm{156}}{\mathrm{3}}\mathrm{arctan}\left(\mathrm{u}\right)+\frac{\mathrm{224u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{152u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{64u}}{\mathrm{3}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{with}\:\mathrm{u}=\mathrm{tan}\frac{\varphi}{\mathrm{2}}\:\mathrm{and}\:\mathrm{x}=\mathrm{2cos}\varphi \\ $$
Commented by mathmax by abdo last updated on 12/Jun/20
$$\mathrm{thankx}\:\mathrm{sir}. \\ $$