Question Number 215974 by ajfour last updated on 23/Jan/25 $$\int\sqrt{\left(\mathrm{2sin}^{−\mathrm{1}} {x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} +{x}^{\mathrm{4}} }{dx} \\ $$ Answered by MathematicalUser2357 last updated on 26/Jan/25 $$\mathrm{No}\:\mathrm{closed}\:\mathrm{forms}.\:\frac{{x}\mid{x}\mid}{\mathrm{2}}+\frac{{x}^{\mathrm{3}} \mid{x}\mid}{\mathrm{3}}−\frac{\mathrm{2}{x}^{\mathrm{5}}…
Question Number 215789 by universe last updated on 18/Jan/25 Commented by mr W last updated on 18/Jan/25 Answered by mr W last updated on 18/Jan/25…
Question Number 215782 by universe last updated on 18/Jan/25 Answered by mr W last updated on 18/Jan/25 Commented by mr W last updated on 18/Jan/25…
Question Number 215775 by ajfour last updated on 17/Jan/25 $$\:\:\underset{\mathrm{0}} {\overset{\:\:{s}} {\int}}\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\left(\mathrm{1}+\sqrt{\mathrm{1}−\left(\frac{{x}}{{s}}\right)^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$ Commented by ajfour last updated on 18/Jan/25 https://youtu.be/ZcHJ_FnQ6Fk?si=r5hLYfUC_lLg_ccq…
Question Number 215754 by universe last updated on 17/Jan/25 Answered by mr W last updated on 17/Jan/25 $${z}=\mathrm{1}+\mathrm{sin}\:\theta\:{with}\:−\frac{\pi}{\mathrm{2}}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${x}={r}\:\mathrm{cos}\:\phi,\:{y}={r}\:\mathrm{sin}\:\phi\:{with}\:\mathrm{0}\leqslant\phi\leqslant\mathrm{2}\pi \\ $$$${I}=\mathrm{2}\pi\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:\theta\int_{\mathrm{0}} ^{\mathrm{cos}\:\theta}…
Question Number 215740 by universe last updated on 16/Jan/25 $$ \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \:\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \:\int_{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} ^{\mathrm{1}+\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}}…
Question Number 215687 by Ajeemkhan last updated on 15/Jan/25 Answered by som(math1967) last updated on 15/Jan/25 $$\int{sec}^{{p}−\mathrm{1}} {xsecxtanxdx} \\ $$$$\Rightarrow\int{sec}^{{p}−\mathrm{1}} {xd}\left({secx}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{p}}{sec}^{{p}} {x}+{C} \\…
Question Number 215704 by universe last updated on 15/Jan/25 Commented by universe last updated on 16/Jan/25 $${HI}\:{MR}\:{W}\:{please}\:{solve}\:{this}\:{problem} \\ $$ Answered by mr W last updated…
Question Number 215604 by universe last updated on 11/Jan/25 Commented by MathematicalUser2357 last updated on 12/Jan/25 $${can}\:{you}\:{please}\:{translate} \\ $$ Answered by mr W last updated…
Question Number 215550 by MrGaster last updated on 10/Jan/25 $$\boldsymbol{\mathrm{Let}}\:\boldsymbol{{u}}^{\left(\mathrm{1}\right)} ,\boldsymbol{{u}}^{\left(\mathrm{2}\right)} \boldsymbol{\mathrm{s}}.\boldsymbol{\mathrm{t}}.\begin{cases}{\boldsymbol{{u}}_{\boldsymbol{{tt}}} ^{\left(\mathrm{1}\right)} =\left(\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial\boldsymbol{{x}}_{{i}} ^{\mathrm{2}} }\right)\boldsymbol{{u}}^{\left(\mathrm{1}\right)} }\\{\boldsymbol{{u}}^{\left(\mathrm{1}\right)} \left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} ,\mathrm{0}\right)=\boldsymbol{\psi}\left(\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}}…