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Question Number 118113 by bemath last updated on 15/Oct/20

$$\int \frac{dx}{x^3\sqrt{x^2-a^2}}=?$$

Answered by Lordose last updated on 15/Oct/20

$$$$$$\mathrm{x}=\mathrm{asecu}\mathrm{dx}=\mathrm{secutanudu}$$$$\frac{1}{\mathrm{a}}\int \frac{\mathrm{secutanu}}{{\sec }^3\mathrm{utanu}}\mathrm{du}$$$$\frac{1}{\mathrm{a}}\int \frac{1}{{\sec }^2\mathrm{u}}\mathrm{du}$$$$\frac{1}{\mathrm{a}}\int {\cos }^2\mathrm{udu}$$$$\frac{1}{\mathrm{a}}(\frac{\mathrm{u}}{2}+\frac{\mathrm{sinucosu}}{2})+\mathrm{C}$$$$\mathrm{secu}=\frac{\mathrm{x}}{\mathrm{a}}\Rightarrow \mathrm{cosu}=\frac{\mathrm{a}}{\mathrm{x}}$$$$\mathrm{sinu}=\sqrt{1-{\cos }^2\mathrm{u}}=\sqrt{1-{\left(\frac{\mathrm{a}}{\mathrm{x}}\right)}^2}$$$$\frac{1}{2\mathrm{a}}({\sec }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\frac{\mathrm{a}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{{\mathrm{x}}^2})+\mathrm{C}$$

Commented by bemath last updated on 15/Oct/20

$$yes...thankyou$$

Answered by 1549442205PVT last updated on 15/Oct/20

$$\mathrm{Put}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}=\mathrm{t}\Rightarrow \mathrm{dt}=\frac{\mathrm{xdx}}{\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}$$$$\mathrm{dx}=\mathrm{tdt}/\mathrm{x}$$$$\int \frac{dx}{x^3\sqrt{x^2-a^2}}=\int \frac{\mathrm{tdt}}{{\mathrm{x}}^4.\mathrm{t}}=\int \frac{\mathrm{dt}}{{({\mathrm{t}}^2+{\mathrm{a}}^2)}^2}$$$${\mathrm{I}}_2=\int \frac{\mathrm{dt}}{{({\mathrm{t}}^2+{\mathrm{a}}^2)}^2}.\mathrm{Put}\mathrm{t}=\mathrm{atan}\phi$$$$\Rightarrow \mathrm{dt}=\mathrm{a}(1+{\tan }^2\phi )\mathrm{d}\phi =\mathrm{a}(1+{\mathrm{t}}^2)\mathrm{d}\phi$$$${\mathrm{I}}_2=\int \frac{\mathrm{a}(1+{\tan }^2\phi )\mathrm{d}\phi }{{\mathrm{a}}^4{({\tan }^2\phi +1)}^2}=\int \frac{\mathrm{d}\phi }{{\mathrm{a}}^3(1+{\tan }^2\phi )}$$$$=\frac{1}{{\mathrm{a}}^3}\int {\cos }^2\phi \mathrm{d}\phi =\frac{1}{{2\mathrm{a}}^3}\int (1+\cos 2\phi )\mathrm{d}\phi$$$$=\frac{\phi }{{2\mathrm{a}}^3}+\frac{1}{{4\mathrm{a}}^3}\sin 2\phi =\frac{{\tan }^{-1}\left(\frac{\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{\mathrm{a}}\right)}{{2\mathrm{a}}^3}$$$$+\frac{1}{{4\mathrm{a}}^3}\times \frac{2\mathrm{a}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{{\mathrm{x}}^2}$$$$=\frac{1}{{2\mathrm{a}}^3}[{\tan }^{-1}\left(\frac{\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{\mathrm{a}}\right)+\frac{\mathrm{a}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}{{\mathrm{x}}^2}]+\mathrm{C}$$

Answered by TANMAY PANACEA last updated on 15/Oct/20

$$t^2=x^2-a^2$$$$\int \frac{xdx}{x^4\sqrt{x^2-a^2}}$$$$\int \frac{tdt}{{(t^2+a^2)}^{2}t}\to t=atan\alpha$$$$\int \frac{{sec}^2\alpha d\alpha }{a^4{sec}^4\alpha }=\frac{1}{a^2}\int \frac{1+cos2\alpha }{2}d\alpha$$$$\frac{1}{a^2}\times \frac{1}{2}(\alpha +\frac{sin2\alpha }{2})+C$$$$\frac{1}{2a^2}({tan}^{-}\left(\frac{t}{a}\right)+\frac{1}{2}\times \frac{2tan\alpha }{1+{tan}^2\alpha })$$$$=\frac{1}{2a^2}{tan}^{-1}\left(\frac{\sqrt{x^2-a^2}}{a}\right)+\frac{1}{2a^2}\left(\frac{\frac{t}{a}}{1+\frac{t^2}{a^2}}\right)+C$$$$=\frac{1}{2a^2}\times {tan}^{-1}\left(\frac{\sqrt{x^2-a^2}}{a}\right)+\frac{1}{2a^2}\left(\frac{\sqrt{x^2-a^2}}{a}\times \frac{a^2}{x^2}\right)+c$$$$=\frac{1}{2a^2}{tan}^{-1}\left(\frac{\sqrt{x^2-a^2}}{a}\right)+\frac{1}{2a^2}\left(\frac{a\sqrt{x^2-a^2}}{x^2}\right)+c$$$$$$$$$$

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