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Question Number 146147 by iloveisrael last updated on 11/Jul/21

$$\mathrm{{\rm Y}}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^4\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}}=?$$

Answered by EDWIN88 last updated on 11/Jul/21

$$Solve\int \frac{dx}{x^4\sqrt{x^2-a^2}}$$$$\mathrm{{\rm Y}}=\int \frac{dx}{x^2.x^3\sqrt{1-a^2{x}^{-2}}}$$$$letu=1-a^2{x}^{-2}\to \frac{du}{2a^2}=\frac{dx}{x^3}$$$$and\frac{a^2}{x^2}=1-u\Rightarrow \frac{1}{x^2}=\frac{1-u}{a^2}$$$$\mathrm{{\rm Y}}=\int \frac{(1-u)du}{2a^4.\sqrt{u}}=\frac{1}{2a^4}\int (u^{-1/2}-u^{1/2})du$$$$\mathrm{{\rm Y}}=\frac{1}{2a^4}(2\sqrt{u}-\frac{2}{3}u\sqrt{u})+c$$$$\mathrm{{\rm Y}}=\frac{\sqrt{u}}{a^4}(1-\frac{1}{3}u)+c$$$$\mathrm{{\rm Y}}=\frac{\sqrt{a^2-x^2}}{a^{4}x}(1-\frac{x^2-a^2}{x^2})+c$$$$\mathrm{{\rm Y}}=\frac{\sqrt{a^2-x^2}}{a^2{x}^3}+c$$

Answered by mathmax by abdo last updated on 11/Jul/21

$$\mathrm{{\rm Y}}{=}_{\mathrm{x}=\frac{\mathrm{a}}{\mathrm{sint}}}\int \frac{{\sin }^4\mathrm{t}}{{\mathrm{a}}^4\sqrt{\frac{{\mathrm{a}}^2}{{\sin }^2\mathrm{t}}-{\mathrm{a}}^2}}(-\frac{\mathrm{acost}}{{\sin }^2\mathrm{t}})\mathrm{dt}$$$$=\frac{1}{{\mathrm{a}}^4}\int \frac{{\sin }^2\mathrm{t}.\mathrm{cost}}{\mathrm{cost}}\mathrm{sint}\mathrm{dt}=\frac{1}{{\mathrm{a}}^4}\int {\sin }^3\mathrm{t}\mathrm{dt}$$$${\sin }^3\mathrm{t}=\mathrm{sint}\left(\frac{1-\cos \left(2\mathrm{t}\right)}{2}\right)=\frac{1}{2}(\mathrm{sint}-\mathrm{sint}\cos \left(2\mathrm{t}\right))$$$$\mathrm{sint}\cos \left(2\mathrm{t}\right)=\cos (\frac{\pi }{2}-\mathrm{t})\cos \left(2\mathrm{t}\right)=\frac{1}{2}\{\cos (\frac{\pi }{2}+\mathrm{t})+\cos (\frac{\pi }{2}-3\mathrm{t}))$$$$=\frac{1}{2}\{-\mathrm{sint}+\sin \left(3\mathrm{t}\right)\}\Rightarrow {\sin }^3\mathrm{t}=\frac{1}{2}\mathrm{sint}-\frac{1}{4}\{-\mathrm{sint}+\sin \left(3\mathrm{t}\right)\}$$$$=\frac{3}{4}\mathrm{sint}-\frac{1}{4}\sin \left(3\mathrm{t}\right)\Rightarrow \mathrm{{\rm Y}}=\frac{1}{{\mathrm{a}}^4}\int (\frac{3}{4}\mathrm{sint}-\frac{1}{4}\sin \left(3\mathrm{t}\right))\mathrm{dt}$$$$=-\frac{3}{{4\mathrm{a}}^4}\mathrm{cost}+\frac{1}{{12\mathrm{a}}^4}\cos \left(3\mathrm{t}\right)+\mathrm{K}$$$$\mathrm{sint}=\frac{\mathrm{a}}{\mathrm{x}}\Rightarrow \mathrm{cost}=\sqrt{1-\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}}$$$$\cos \left(3\mathrm{t}\right)=\mathrm{cost}.\cos 2\mathrm{t}-\mathrm{sint}\sin \left(2\mathrm{t}\right)$$$$=\mathrm{cost}({2\cos }^2\mathrm{t}-1)-{2\sin }^2\mathrm{t}\mathrm{cost}$$$$=\sqrt{1-\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}}(2(1-\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2})-1)-2\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}\sqrt{1-\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}}\Rightarrow$$$$\mathrm{\Psi }=-\frac{3}{{4\mathrm{a}}^4}\sqrt{1-\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}}+\frac{1}{{12\mathrm{a}}^4}\sqrt{1-\frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}}(1-\frac{{4\mathrm{a}}^2}{{\mathrm{x}}^2})+\mathrm{K}$$

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