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Question Number 142457 by ajfour last updated on 01/Jun/21

$$I=\int \frac{dx}{\sqrt{a^2-(x+\frac{1}{x})}}$$

Answered by MJS_new last updated on 01/Jun/21

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{afraid}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{ending}\mathrm{on}\mathrm{elliptic}\mathrm{functions}$$$$\mathrm{again}.$$$$\mathrm{my}\mathrm{try}:$$$$\int \frac{dx}{\sqrt{a^2-(x+\frac{1}{x})}}=$$$$[t=x+\frac{1}{x}\to dx=\frac{t+\sqrt{t^2-4}}{\sqrt{t^2-4}}]$$$$=\frac{1}{2}\int \frac{t+\sqrt{t^2-4}}{\sqrt{t^2-4}\sqrt{a^2-t}}dt=$$$$=\frac{1}{2}\int \frac{dt}{\sqrt{a^2-t}}+\frac{1}{2}\int \frac{t}{\sqrt{t^2-4}\sqrt{a^2-t}}dt=$$$$=-\sqrt{a^2-t}+\frac{1}{2}\int \frac{t}{\sqrt{t^2-4}\sqrt{a^2-t}}dt$$$$\frac{1}{2}\int \frac{t}{\sqrt{t^2-4}\sqrt{a^2-t}}dt=$$$$[u=\frac{\sqrt{t^2-4}}{\sqrt{a^2-t}}\to dt=-\frac{2\sqrt{t^2-4}\sqrt{{(a^2-t)}^3}}{t^2-2a^{2}t+4}]$$$$=\frac{1}{2}\int du-\frac{1}{2}\int \frac{u^2}{\sqrt{u^4+4a^2{u}^2+16}}du=$$$$=\frac{u}{2}-\frac{1}{2}\int \frac{u^2}{\sqrt{u^4+4a^2{u}^2+16}}du$$$$\mathrm{and}\mathrm{we}\mathrm{cannot}\mathrm{solve}\mathrm{the}\mathrm{remaining}\mathrm{integral}$$$$\mathrm{using}\mathrm{elementary}\mathrm{calculus}$$

Commented by ajfour last updated on 01/Jun/21

$$I=\int \frac{u^{2}du}{\sqrt{{(u^2+2a^2)}^2-4(a^4-4)}}$$$$I=\int \frac{du}{\sqrt{{(1+\frac{2a^2}{u^2})}^2-{\left(\frac{s}{u^2}\right)}^2}}$$$$let\frac{2a^2}{u^2}=\cos 2\theta$$$$u^2=2a^2\sec 2\theta$$$$2udu=4a^2\sec 2\theta \sqrt{\sec {}^{2}2\theta -1}d\theta$$$$du=\sqrt{2}a\sec 2\theta \sqrt{\sec 2\theta -\cos 2\theta }$$$$I=\int \frac{\sqrt{2}a\sec 2\theta \sqrt{\sec 2\theta -\cos 2\theta }d\theta }{\sqrt{4\cos {}^4\theta -\frac{s^2}{4a^4}\cos {}^{2}2\theta }}$$$$I=\int \frac{\sqrt{2}a\sin 2\theta d\theta }{\sqrt{{(1+\cos 2\theta )}^2\cos 2\theta -\frac{s^2}{4a^4}\cos {}^{3}2\theta }}$$$$I=-\frac{a}{\sqrt{2}}\int \frac{dz}{\sqrt{z{(1+z)}^2-m^2{z}^3}}$$$$...$$

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