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Author: Tinku Tara

x-2-y-2-4-x-2-y-2-5-find-y-x-

Question Number 201302 by hardmath last updated on 03/Dec/23 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{−\mathrm{2}} \:=\:\mathrm{4}}\\{\mathrm{x}^{−\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{5}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:? \\ $$ Answered by witcher3 last updated on 03/Dec/23 $$\left(\mathrm{1}\right)\ast\left(\mathrm{2}\right)\Leftrightarrow\mathrm{2}+\left(\mathrm{xy}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\left(\mathrm{xy}\right)^{\mathrm{2}}…

Question-201292

Question Number 201292 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\underset{−{a}} {\overset{{a}} {\int}}\frac{{cos}\left({x}\right){dx}}{\mathrm{1}+{e}^{\pi/{x}} }=\underset{{a}} {\overset{−{a}} {\int}}\frac{{cos}\left(−{x}\right){d}\left(−{x}\right)}{\mathrm{1}+{e}^{\pi/\left(−{x}\right)} }= \\ $$$$=\int_{−{a}}…

Question-201293

Question Number 201293 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right){dx}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}= \\ $$$$=\int_{\mathrm{1}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left(\left(\mathrm{2}{x}\right)/\mathrm{2}\right)}{\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{4}\left(\mathrm{2}{x}\right)}{d}\left(\mathrm{2}{x}\right)=\mathrm{2}\int_{\mathrm{1}}…

Question-201290

Question Number 201290 by sonukgindia last updated on 03/Dec/23 Answered by Calculusboy last updated on 03/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{By}}\:\boldsymbol{{using}}\:\boldsymbol{{kings}}\:\boldsymbol{{rule}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}}…

Question-201291

Question Number 201291 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${notice} \\ $$$${x}=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}},\:{dx}=−\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}…

Question-201282

Question Number 201282 by Supro last updated on 03/Dec/23 Answered by MM42 last updated on 03/Dec/23 $${x}=\frac{\begin{vmatrix}{{a}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:{b}}\\{{ab}\:\:\:\:\:\:−{a}}\end{vmatrix}}{\begin{vmatrix}{{a}\:\:\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:−{a}}\end{vmatrix}}=\frac{−{a}^{\mathrm{3}} −{ab}^{\mathrm{2}} }{−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:={a} \\ $$$${y}=\frac{\begin{vmatrix}{{a}\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} }\\{{b}\:\:\:\:\:\:\:\:{ab}}\end{vmatrix}}{\begin{vmatrix}{{a}\:\:\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:−{a}}\end{vmatrix}}=\frac{{a}^{\mathrm{2}}…