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Question Number 203836 by patrice last updated on 29/Jan/24

Answered by Frix last updated on 30/Jan/24

$$\mathrm{In}3\mathrm{steps}:$$$$1.t=x^2-1$$$$2.u=\frac{5t+\sqrt{25t^2+11}}{11}$$$$3.v=\frac{\sqrt{1+u}}{\sqrt{1-u}}\Rightarrow$$$$\frac{4}{5}\underset{\frac{\sqrt{6}}{\sqrt{5}}}{\stackrel{\mathrm{\infty }}{\int }}\frac{v^2(3v^4-5v^2+3)}{{(v^4-1)}^2}dv=$$$$={[\frac{v(5v^2-6)}{5(v^4-1)}+\frac{\ln \frac{v+1}{v-1}}{20}+\frac{{11\tan }^{-1}v}{10}]}_{\frac{\sqrt{6}}{\sqrt{5}}}^{\mathrm{\infty }}=$$$$=\frac{11\pi }{20}-\frac{\ln (\sqrt{6}+\sqrt{5})}{10}-\frac{{11\tan }^{-1}\frac{\sqrt{6}}{\sqrt{5}}}{10}$$

Commented by patrice last updated on 30/Jan/24

je ne sais pas bien merci

Answered by Sutrisno last updated on 30/Jan/24

$$$$$$={\int }_0^{\sqrt{2}}\frac{\sqrt{6-{(5x^2-6)}^2}}{\sqrt{5}}dx$$$$={\int }_0^{\sqrt{2}}\frac{\sqrt{12-5x^2}}{\sqrt{5}}dx$$$$misal\sqrt{5}x=\sqrt{12}sin\theta \to \sqrt{5}dx=\sqrt{12}cos\theta d\theta$$$$=\int \frac{\sqrt{12-12{sin}^2\theta }}{\sqrt{5}}.\frac{\sqrt{12}cos\theta d\theta }{\sqrt{5}}$$$$=\frac{12}{5}\int {cos}^2\theta d\theta$$$$=\frac{12}{5}\int \frac{cos2\theta +1}{2}d\theta$$$$=\frac{6}{5}(\frac{1}{2}sin2\theta +\theta )$$$$=\frac{6}{5}(sin\theta cos\theta +\theta )$$$$=\frac{6}{5}(\frac{\sqrt{5}x}{\sqrt{12}}.\frac{\sqrt{12-5x^2}}{\sqrt{12}}+{sin}^{-1}\left(\frac{\sqrt{5}x}{\sqrt{12}}\right))\underset{0}{\stackrel{\sqrt{2}}{\mid }}$$$$=\frac{6}{5}(\frac{\sqrt{10}}{\sqrt{12}}.\frac{\sqrt{2}}{\sqrt{12}}+{sin}^{-1}\left(\frac{\sqrt{10}}{\sqrt{12}}\right))$$$$=\frac{6}{5}(\frac{\sqrt{5}}{6}+{sin}^{-1}\left(\frac{\sqrt{30}}{6}\right))$$$$$$

Commented by Frix last updated on 30/Jan/24

$$\mathrm{Wrong}\mathrm{because}$$$$\sqrt{25x^4-50x^2+36}\ne 5x^2-6$$$$\mathrm{And}\mathrm{we}\mathrm{see}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{a}\mathrm{typo}\mathrm{because}\mathrm{of}\mathrm{the}$$$$\mathrm{upper}\mathrm{border}\sqrt{2}\mathrm{where}\mathrm{the}\mathrm{given}\mathrm{function}$$$$\mathrm{equals}\mathrm{zero}.$$

Commented by Frix last updated on 30/Jan/24

$$\underset{0}{\stackrel{b}{\int }}\frac{\sqrt{6-\sqrt{25x^4-60x^2+36}}}{\sqrt{5}}dx=$$$$=\underset{0}{\stackrel{b}{\int }}\frac{\sqrt{6-\mid 5x^2-6\mid }}{\sqrt{5}}dx=$$$$=\underset{0}{\stackrel{\frac{\sqrt{30}}{5}}{\int }}xdx+\underset{\frac{\sqrt{30}}{5}}{\stackrel{b}{\int }}\frac{\sqrt{12-5x^2}}{\sqrt{5}}dx=$$$$={\left[\frac{x^2}{2}\right]}_0^{\frac{\sqrt{30}}{5}}+{[\frac{x\sqrt{12-5x^2}}{2\sqrt{5}}+\frac{{6\sin }^{-1}\frac{\sqrt{15}x}{6}}{5}]}_{\frac{\sqrt{30}}{5}}^b$$$$\mathrm{This}\mathrm{gives}:$$$$b=\sqrt{2}$$$$\frac{\sqrt{5}}{5}-\frac{3\pi }{10}+\frac{{6\sin }^{-1}\frac{\sqrt{30}}{6}}{5}$$$$b=\frac{2\sqrt{15}}{5}\left[\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{function}\right]$$$$\frac{3\pi }{10}$$

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