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Author: Tinku Tara

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}}…

A-ball-lies-on-the-function-z-xy-at-the-point-1-2-2-Find-the-point-in-the-xy-plane-where-the-ball-will-touch-it-an-unsolved-old-question-Q200929-

Question Number 201214 by mr W last updated on 02/Dec/23 $$\mathrm{A}\:\mathrm{ball}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{function}\:{z}={xy}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{the}\:{xy}−\mathrm{plane}\:\mathrm{where}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{will} \\ $$$$\mathrm{touch}\:\mathrm{it}. \\ $$$$ \\ $$$$\left({an}\:{unsolved}\:{old}\:{question}\:{Q}\mathrm{200929}\right) \\ $$ Answered by…