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Author: Tinku Tara

Prove-0-1-K-2-36K-2-6K-K-2-36K-2-dk-k-1-k-2-2-3-pi-2-4-2-2-3-0-1-K-2-36K-2-6K-2-K-2-36K-2-dk-k-1-k-2-2-3-

Question Number 222567 by MrGaster last updated on 30/Jun/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}^{'} }{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }\:}\frac{{dk}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\sqrt{\pi}\left(\sqrt{\mathrm{2}}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\right) \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}}…

n-1-1-cosh-4-n-ln-2-1-

Question Number 222592 by Tawa11 last updated on 30/Jun/25 $$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{cosh}^{\mathrm{4}} \left[\mathrm{n}\:\mathrm{ln}\left(\sqrt{\mathrm{2}}\:\:+\:\:\mathrm{1}\right)\right]} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

x-x-64-x-

Question Number 222552 by mathlove last updated on 29/Jun/25 $${x}^{{x}} =\mathrm{64}\:\:\:\:\:\:,\:\:{x}=? \\ $$ Answered by mr W last updated on 29/Jun/25 $${x}\mathrm{ln}\:{x}=\mathrm{ln}\:\mathrm{64} \\ $$$$\mathrm{ln}\:{xe}^{\mathrm{ln}\:{x}} =\mathrm{6ln}\:\mathrm{2}…

Question-222520

Question Number 222520 by mr W last updated on 29/Jun/25 Commented by mr W last updated on 29/Jun/25 $${a}\:{ball}\:{is}\:{released}\:{at}\:{the}\:{point}\:\left({d},\:{h}\right) \\ $$$${on}\:{to}\:{the}\:{ground}\:\:{which}\:{has}\:{the}\: \\ $$$${shape}\:{of}\:{a}\:{hyperbola} \\ $$$${with}\:{equation}\:−\frac{{x}^{\mathrm{2}}…

x-V-J-x-ustx-zac-2-x-x-ustx-0-x-1-v-pi-2-L-l-x-0-v-X-2x-2-1-

Question Number 222523 by hu last updated on 29/Jun/25 $${x}+{V}−{J}\left({x}\right)\frac{{ustx}}{{zac}^{\mathrm{2}} {x}}={x}−\frac{{ustx}_{\mathrm{0}} }{\:\sqrt{{x}}\psi+\zeta\left(\frac{−\mathrm{1}+{v\%}}{\pi+\mathrm{2}}+\mathscr{L}_{{l}\left({x}\rightarrow\mathrm{0}\right)} ^{\:\:{v\%}} {X}_{\mathrm{2}{x}^{\mathrm{2}} } ^{\:\mathrm{1}} \right)} \\ $$ Answered by wewji12 last updated on…

Given-the-integer-k-how-to-find-the-incomplete-general-solution-for-the-non-trivial-integer-solutions-of-the-Diophantine-equation-a-4-b-4-ka-2-b-2-c-4-d-4-kc-2-d-2-a-b-c-d-N-k-Z-gcd-a-b-c-d-

Question Number 222516 by MrGaster last updated on 29/Jun/25 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{integer}\:{k},\mathrm{how}\:\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{incomplete}\:\mathrm{general} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\mathrm{the}\:\mathrm{non}-\mathrm{trivial}\:\mathrm{integer} \\ $$$$\mathrm{solutions}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{Diophantine}\:\mathrm{equation}: \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{ka}^{\mathrm{2}} {b}^{\mathrm{2}} ={c}^{\mathrm{4}} +{d}^{\mathrm{4}} +{kc}^{\mathrm{2}} {d}^{\mathrm{2}}…

very-very-crazy-problem-i-am-not-found-what-is-the-result-of-this-integral-1-z-z-h-z-2h-dz-

Question Number 222541 by Nicholas666 last updated on 29/Jun/25 $$ \\ $$$$\:\:\:\mathrm{very}\:\mathrm{very}\:\mathrm{crazy}\:\mathrm{problem},\:\:\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{found} \\ $$$$\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{result}\:\mathrm{of}\:\mathrm{this}\:\mathrm{integral}; \\ $$$$\:\:\:\:\:\:\:\:\int\frac{\mathrm{1}}{\:\sqrt{{z}}\:+\:\sqrt{{z}−{h}}\:+\:\sqrt{{z}−\mathrm{2}{h}}}\:{dz} \\ $$$$ \\ $$ Commented by Ghisom last updated…