Question Number 212067 by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Commented by Ghisom last updated on 28/Sep/24 $$\mathrm{that}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\…
Question Number 212024 by Spillover last updated on 27/Sep/24 Answered by Frix last updated on 27/Sep/24 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\left(\mathrm{tan}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} {dx}\:\overset{\left[{t}=\left(\mathrm{tan}\:{x}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \right]} {=}\:\mathrm{3}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{6}} +\mathrm{1}}…
Question Number 212033 by RojaTaniya last updated on 27/Sep/24 $$\:{Find}\:{last}\:{two}\:{digit}\:{of}\:\mid\mathrm{33}^{\mathrm{22}} −\mathrm{22}^{\mathrm{33}} \mid \\ $$ Answered by Frix last updated on 27/Sep/24 $$\mathrm{22}^{\mathrm{33}} −\mathrm{33}^{\mathrm{22}} = \\…
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Question Number 211989 by Spillover last updated on 26/Sep/24 Answered by som(math1967) last updated on 26/Sep/24 $$\:{cos}\alpha{cos}\beta+{sin}\alpha{sin}\beta+{cos}\beta{cos}\gamma \\ $$$$+{sin}\beta{sin}\gamma+{cos}\gamma{cos}\alpha+{sin}\gamma{sin}\alpha \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}\left({cos}\alpha{cos}\beta+{cos}\beta{cos}\gamma+{cos}\gamma{cos}\alpha\right) \\ $$$$+\mathrm{2}\left({sin}\alpha{sin}\beta+{sin}\beta{sin}\gamma+{sin}\alpha{sin}\gamma\right)…
Question Number 211987 by Spillover last updated on 26/Sep/24 Commented by MathematicalUser2357 last updated on 26/Sep/24 $$\mathrm{No}\:\mathrm{closed}\:\mathrm{forms}\:\mathrm{for}\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:{x}}}−\sqrt{\mathrm{tan}\:{x}}\right){dx}. \\ $$$$\mathrm{But}\:\mathrm{you}\:\mathrm{can}\:\mathrm{just}\:\mathrm{approximate}. \\ $$ Commented…
Question Number 211986 by Spillover last updated on 26/Sep/24 Answered by Spillover last updated on 28/Sep/24 Answered by Spillover last updated on 28/Sep/24 Commented by…
Question Number 212006 by RojaTaniya last updated on 26/Sep/24 Answered by a.lgnaoui last updated on 26/Sep/24 $$\:\:\mathrm{posons}\:\:\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{x}}+\mathrm{100} \\ $$$$\:\:\frac{\left(\boldsymbol{\mathrm{z}}−\mathrm{2}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{2}\right)^{\mathrm{5}} }{\left(\boldsymbol{\mathrm{z}}−\mathrm{1}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}} +\mathrm{80}\boldsymbol{\mathrm{z}}^{\mathrm{3}} +\mathrm{160}\boldsymbol{\mathrm{z}}}{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}}…
Question Number 212002 by universe last updated on 26/Sep/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211949 by Spillover last updated on 25/Sep/24 Answered by MathematicalUser2357 last updated on 26/Sep/24 $$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{idea}! \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\mathrm{ln}\left(\frac{{ax}^{\mathrm{2}} +\sqrt{\mathrm{2}{a}}{x}+\mathrm{1}}{{ax}^{\mathrm{2}} −\sqrt{\mathrm{2}{a}}{x}+\mathrm{1}}\right){dx}=\mathrm{4}\pi\:\mathrm{cot}^{−\mathrm{1}} \sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}}{{a}}}…