Question Number 118113 by bemath last updated on 15/Oct/20
$$\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:=?\: \\ $$
Answered by Lordose last updated on 15/Oct/20
$$ \\ $$$$\mathrm{x}=\mathrm{asecu}\:\mathrm{dx}=\:\mathrm{secutanudu} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\int\:\frac{\mathrm{secutanu}}{\mathrm{sec}^{\mathrm{3}} \mathrm{utanu}}\mathrm{du} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\int\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \mathrm{u}}\mathrm{du} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\int\mathrm{cos}^{\mathrm{2}} \mathrm{udu} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\left(\frac{\mathrm{u}}{\mathrm{2}}\:+\:\frac{\mathrm{sinucosu}}{\mathrm{2}}\right)\:+\:\mathrm{C} \\ $$$$\mathrm{secu}=\:\frac{\mathrm{x}}{\mathrm{a}}\:\Rightarrow\:\mathrm{cosu}=\frac{\mathrm{a}}{\mathrm{x}} \\ $$$$\mathrm{sinu}=\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{u}}\:=\:\sqrt{\mathrm{1}−\left(\frac{\mathrm{a}}{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2a}}\left(\mathrm{sec}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right)\:+\:\frac{\mathrm{a}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} }\right)\:+\:\mathrm{C} \\ $$
Commented by bemath last updated on 15/Oct/20
$${yes}...{thank}\:{you} \\ $$
Answered by 1549442205PVT last updated on 15/Oct/20
![Put (√(x^2 −a^2 )) =t⇒dt=((xdx)/( (√(x^2 −a^2 )))) dx=tdt/x ∫ (dx/(x^3 (√(x^2 −a^2 )))) =∫ ((tdt)/(x^4 .t))=∫(dt/((t^2 +a^2 )^2 )) I_2 =∫(dt/((t^2 +a^2 )^2 )).Put t=atanϕ ⇒dt=a(1+tan^2 ϕ)dϕ=a(1+t^2 )dϕ I_2 =∫((a(1+tan^2 ϕ)dϕ)/(a^4 (tan^2 ϕ+1)^2 ))=∫(dϕ/(a^3 (1+tan^2 ϕ))) =(1/a^3 )∫cos^2 ϕdϕ=(1/(2a^3 ))∫(1+cos2ϕ)dϕ =(ϕ/(2a^3 ))+(1/(4a^3 ))sin2ϕ=((tan^(−1) (((√(x^2 −a^2 ))/a)))/(2a^3 )) +(1/(4a^3 ))×((2a(√(x^2 −a^2 )))/x^2 ) =(1/(2a^3 ))[tan^(−1) (((√(x^2 −a^2 ))/a))+((a(√(x^2 −a^2 )))/x^2 )]+C](Q118126.png)
$$\mathrm{Put}\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\:=\mathrm{t}\Rightarrow\mathrm{dt}=\frac{\mathrm{xdx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }} \\ $$$$\mathrm{dx}=\mathrm{tdt}/\mathrm{x} \\ $$$$\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:=\int\:\frac{\mathrm{tdt}}{\mathrm{x}^{\mathrm{4}} .\mathrm{t}}=\int\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{I}_{\mathrm{2}} =\int\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }.\mathrm{Put}\:\mathrm{t}=\mathrm{atan}\varphi \\ $$$$\Rightarrow\mathrm{dt}=\mathrm{a}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \varphi\right)\mathrm{d}\varphi=\mathrm{a}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{d}\varphi \\ $$$$\mathrm{I}_{\mathrm{2}} =\int\frac{\mathrm{a}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \varphi\right)\mathrm{d}\varphi}{\mathrm{a}^{\mathrm{4}} \left(\mathrm{tan}^{\mathrm{2}} \varphi+\mathrm{1}\right)^{\mathrm{2}} }=\int\frac{\mathrm{d}\varphi}{\mathrm{a}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \varphi\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }\int\mathrm{cos}^{\mathrm{2}} \varphi\mathrm{d}\varphi=\frac{\mathrm{1}}{\mathrm{2a}^{\mathrm{3}} }\int\left(\mathrm{1}+\mathrm{cos2}\varphi\right)\mathrm{d}\varphi \\ $$$$=\frac{\varphi}{\mathrm{2a}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4a}^{\mathrm{3}} }\mathrm{sin2}\varphi=\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}}\right)}{\mathrm{2a}^{\mathrm{3}} } \\ $$$$+\frac{\mathrm{1}}{\mathrm{4a}^{\mathrm{3}} }×\frac{\mathrm{2a}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2a}^{\mathrm{3}} }\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}}\right)+\frac{\mathrm{a}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} }\right]+\mathrm{C} \\ $$
Answered by TANMAY PANACEA last updated on 15/Oct/20
$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\int\frac{{xdx}}{{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$$\int\frac{{tdt}}{\left({t}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} {t}}\rightarrow{t}={atan}\alpha \\ $$$$\int\frac{{sec}^{\mathrm{2}} \alpha{d}\alpha}{{a}^{\mathrm{4}} {sec}^{\mathrm{4}} \alpha}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}}{d}\alpha \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha+\frac{{sin}\mathrm{2}\alpha}{\mathrm{2}}\right)+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\left({tan}^{−} \left(\frac{{t}}{{a}}\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{tan}\alpha}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}\right)+\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\left(\frac{\frac{{t}}{{a}}}{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }×{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}\right)+\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\left(\frac{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}×\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}\:\right)+\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\left(\frac{{a}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }\right)+{c} \\ $$$$ \\ $$$$ \\ $$