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Question Number 213369 by Frix last updated on 03/Nov/24

$$\mathrm{Old}\mathrm{question}203835$$$$\underset{0}{\stackrel{\sqrt{2}}{\int }}\frac{\sqrt{6-\sqrt{25x^4-50x^2+36}}}{\sqrt{5}}dx=?$$

Answered by Frix last updated on 03/Nov/24

$$\underset{0}{\stackrel{\sqrt{2}}{\int }}\frac{\sqrt{6-\sqrt{25x^4-50x^2+36}}}{\sqrt{5}}dx\stackrel{[t=\frac{5}{\sqrt{11}}(x^2-1)]}{=}$$$$=\frac{\sqrt{11}}{10}\underset{-\frac{5}{\sqrt{11}}}{\stackrel{\frac{5}{\sqrt{11}}}{\int }}\frac{\sqrt{6-\sqrt{11(t^2+1)}}}{\sqrt{\sqrt{11}t+5}}dt\stackrel{[u=t+\sqrt{t^2+1}]}{=}$$$$=\frac{\sqrt{11}}{20}\underset{\frac{1}{\sqrt{11}}}{\stackrel{\sqrt{11}}{\int }}\frac{(u^2+1)\sqrt{-\sqrt{11}{u}^2+12u-\sqrt{11}}}{u^2\sqrt{\sqrt{11}{u}^2+10u-\sqrt{11}}}du=$$$$=\frac{\sqrt{11}}{20}\underset{\frac{1}{\sqrt{11}}}{\stackrel{\sqrt{11}}{\int }}\frac{(u^2+1)\sqrt{(\sqrt{11}-u)(u-\frac{1}{\sqrt{11}})}}{u^2\sqrt{(\sqrt{11}+u)(u-\frac{1}{\sqrt{11}})}}du=$$$$=\frac{\sqrt{11}}{20}\underset{\frac{1}{\sqrt{11}}}{\stackrel{\sqrt{11}}{\int }}\frac{(u^2+1)\sqrt{\sqrt{11}-u}}{u^2\sqrt{\sqrt{11}+u}}du\stackrel{[v=\frac{\sqrt{\sqrt{11}-u}}{\sqrt{\sqrt{11}+u}}]}{=}$$$$=\frac{4}{5}\underset{0}{\stackrel{\frac{\sqrt{5}}{\sqrt{6}}}{\int }}\frac{v^2(3v^4-5v^2+3)}{{(v-1)}^2{(v+1)}^2{(v^2+1)}^2}dv=$$$$=\frac{1}{20}\underset{0}{\stackrel{\frac{\sqrt{5}}{\sqrt{6}}}{\int }}(\frac{1}{{(v-1)}^2}+\frac{1}{{(v+1)}^2}+\frac{1}{v-1}-\frac{1}{v+1})dv+$$$$+\frac{11}{5}\underset{0}{\stackrel{\frac{\sqrt{5}}{\sqrt{6}}}{\int }}(\frac{1}{v^2+1}-\frac{1}{{(v^2+1)}^2})dv=$$$$...$$$$={[-\frac{v(6v^2-5)}{5(v^4-1)}+\frac{1}{20}\ln \mid \frac{v-1}{v+1}\mid +\frac{11}{10}{\tan }^{-1}v]}_0^{\frac{\sqrt{5}}{\sqrt{6}}}=$$$$=\frac{1}{10}\ln (\sqrt{6}-\sqrt{5})+\frac{11}{10}{\tan }^{-1}\frac{\sqrt{5}}{\sqrt{6}}$$

Answered by MrGaster last updated on 03/Nov/24

$${\int }_0^{\sqrt{2}}\frac{\sqrt{6-\sqrt{{(5x^2-6)}^2}}}{\sqrt{5}}dx={\int }_0^{\sqrt{2}}\frac{\sqrt{6-\mid 5x^2-6\mid }}{\sqrt{5}}dx$$$$={\int }_0^{\sqrt{2}}\frac{\sqrt{6-(6-5x^2)}}{\sqrt{5}}dx\mathrm{for}0\le x\le \sqrt{2},5x^2-6\le 0$$$$={\int }_0^{\sqrt{2}}xdx=\frac{x^2}{2}{\mid }_0^{\sqrt{2}}$$$$=\frac{{\left(\sqrt{2}\right)}^2}{2}-\frac{0^2}{2}$$$$=\frac{2}{1}$$$$=1\mathrm{or}\approx 0.65942035657438469264675199091487656785970133168787978938786781873469280290781916555575665168451357245368318166250688730704685276141919488865053159337...$$

Commented by Frix last updated on 04/Nov/24

$$\mathrm{Again}:\mathrm{this}\mathrm{is}\mathrm{wrong}$$$$25x^4-50x^2+36\ne {(5x^2-6)}^2=25x^4-60x^2+36$$$$\mathrm{And}\frac{2}{1}\ne 1\ne .0659...$$$$\mathrm{I}\mathrm{doubt}\mathrm{you}\mathrm{have}\mathrm{any}\mathrm{clue}\mathrm{at}\mathrm{all}$$

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