Question Number 201298 by ajfour last updated on 03/Dec/23 Commented by ajfour last updated on 03/Dec/23 $${Find}\:{IJ}\:\:\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$ Commented by mr W last updated…
Question Number 201292 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\underset{−{a}} {\overset{{a}} {\int}}\frac{{cos}\left({x}\right){dx}}{\mathrm{1}+{e}^{\pi/{x}} }=\underset{{a}} {\overset{−{a}} {\int}}\frac{{cos}\left(−{x}\right){d}\left(−{x}\right)}{\mathrm{1}+{e}^{\pi/\left(−{x}\right)} }= \\ $$$$=\int_{−{a}}…
Question Number 201293 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\int_{\mathrm{2}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left({x}/\mathrm{2}\right){dx}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}= \\ $$$$=\int_{\mathrm{1}} ^{\:\infty} \frac{\mathrm{8}{arcsec}\left(\left(\mathrm{2}{x}\right)/\mathrm{2}\right)}{\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{4}\left(\mathrm{2}{x}\right)}{d}\left(\mathrm{2}{x}\right)=\mathrm{2}\int_{\mathrm{1}}…
Question Number 201289 by sonukgindia last updated on 03/Dec/23 Commented by mr W last updated on 03/Dec/23 $${sir}: \\ $$$${why}\:{do}\:{you}\:{ignore}\:{all}\:{questions} \\ $$$${other}\:{people}\:{are}\:{asking}\:{you}? \\ $$ Answered…
Question Number 201290 by sonukgindia last updated on 03/Dec/23 Answered by Calculusboy last updated on 03/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{By}}\:\boldsymbol{{using}}\:\boldsymbol{{kings}}\:\boldsymbol{{rule}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}{\boldsymbol{{sin}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)+\boldsymbol{{cos}}^{\boldsymbol{\varphi}} \left(\boldsymbol{{x}}\right)}\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}}…
Question Number 201291 by sonukgindia last updated on 03/Dec/23 Answered by aleks041103 last updated on 03/Dec/23 $${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${notice} \\ $$$${x}=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}},\:{dx}=−\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}…
Question Number 201286 by sonukgindia last updated on 03/Dec/23 Commented by mr W last updated on 03/Dec/23 $${no}\:{other}\:{data}?\:{i}\:{don}'{t}\:{think}\:{it}'{s} \\ $$$${possible}\:{to}\:{solve}\:{with}\:{given}\:{data}. \\ $$ Terms of Service…
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Question Number 201282 by Supro last updated on 03/Dec/23 Answered by MM42 last updated on 03/Dec/23 $${x}=\frac{\begin{vmatrix}{{a}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:{b}}\\{{ab}\:\:\:\:\:\:−{a}}\end{vmatrix}}{\begin{vmatrix}{{a}\:\:\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:−{a}}\end{vmatrix}}=\frac{−{a}^{\mathrm{3}} −{ab}^{\mathrm{2}} }{−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:={a} \\ $$$${y}=\frac{\begin{vmatrix}{{a}\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} }\\{{b}\:\:\:\:\:\:\:\:{ab}}\end{vmatrix}}{\begin{vmatrix}{{a}\:\:\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:−{a}}\end{vmatrix}}=\frac{{a}^{\mathrm{2}}…
Question Number 201214 by mr W last updated on 02/Dec/23 $$\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…