Question Number 83146 by 09658867628 last updated on 28/Feb/20 $$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\:\frac{\sqrt{\mathrm{cot}\:{x}}}{\:\sqrt{\mathrm{cot}\:{x}}\:+\:\sqrt{\mathrm{tan}\:{x}}}\:{dx}\:= \\ $$ Answered by Kunal12588 last updated on 28/Feb/20 $${I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \frac{\sqrt{{cot}\:{x}}}{\:\sqrt{{cot}\:{x}}+\sqrt{{tan}\:{x}}}{dx} \\…
Question Number 83145 by 09658867628 last updated on 28/Feb/20 $$\:\underset{\:\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:{e}^{\sqrt{{x}}} \:{dx}\:= \\ $$ Commented by mathmax by abdo last updated on 28/Feb/20 $${I}=\int_{\mathrm{1}}…
Question Number 83078 by mhmd last updated on 27/Feb/20 $$\mathrm{If}\:\:\:{I}_{\mathrm{1}} =\underset{{e}} {\overset{{e}^{\mathrm{2}} } {\int}}\:\frac{{dx}}{\mathrm{log}\:{x}}\:\:\mathrm{and}\:\:{I}_{\mathrm{2}} =\:\underset{\:\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{{e}^{{x}} }{{x}}\:{dx},\:\mathrm{then} \\ $$ Commented by mathmax by abdo…
Question Number 83077 by mhmd last updated on 27/Feb/20 $$\mathrm{If}\:\:\:{I}_{\mathrm{1}} =\underset{{e}} {\overset{{e}^{\mathrm{2}} } {\int}}\:\frac{{dx}}{\mathrm{log}\:{x}}\:\:\mathrm{and}\:\:{I}_{\mathrm{2}} =\:\underset{\:\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\frac{{e}^{{x}} }{{x}}\:{dx},\:\mathrm{then} \\ $$ Answered by TANMAY PANACEA last…
Question Number 83075 by mhmd last updated on 27/Feb/20 $$\underset{−\mathrm{1}\:\:} {\overset{\mathrm{3}} {\int}}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\right){dx}\:= \\ $$ Commented by mathmax by abdo last updated…
Question Number 82540 by zainal tanjung last updated on 22/Feb/20 $$\mathrm{The}\:\mathrm{arithmetic}\:\mathrm{mean}\:\mathrm{between}\:\mathrm{two}\:\mathrm{numbers} \\ $$$$\mathrm{is}\:{A},\:\mathrm{and}\:{S}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:{n}\:\mathrm{arithmetic}\:\mathrm{means} \\ $$$$\mathrm{between}\:\mathrm{these}\:\mathrm{numbers},\:\mathrm{then} \\ $$ Answered by TANMAY PANACEA last updated on 22/Feb/20…
Question Number 16983 by aamaguestt last updated on 29/Jun/17 $$\mathrm{If}\:\mathrm{15}\:\mathrm{men}\:\mathrm{or}\:\mathrm{24}\:\mathrm{women}\:\mathrm{or}\:\mathrm{36}\:\mathrm{boys}\:\mathrm{do}\:\mathrm{a}\: \\ $$$$\mathrm{piece}\:\mathrm{of}\:\mathrm{work}\:\mathrm{in}\:\mathrm{12}\:\mathrm{days},\:\mathrm{working}\:\mathrm{8}\:\mathrm{hours} \\ $$$$\mathrm{a}\:\mathrm{day},\:\mathrm{how}\:\mathrm{many}\:\mathrm{men}\:\mathrm{must}\:\mathrm{associated} \\ $$$$\mathrm{with}\:\mathrm{12}\:\mathrm{women}\:\mathrm{and}\:\mathrm{6}\:\mathrm{boys}\:\mathrm{to}\:\mathrm{do}\:\mathrm{another} \\ $$$$\mathrm{piece}\:\mathrm{of}\:\mathrm{work}\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{times}\:\mathrm{as}\:\mathrm{great}\:\mathrm{in}\:\mathrm{30} \\ $$$$\mathrm{days}\:\mathrm{working}\:\mathrm{6}\:\mathrm{hrs}\:\mathrm{a}\:\mathrm{day}? \\ $$ Answered by ajfour…
Question Number 82474 by zainal tanjung last updated on 21/Feb/20 $$\underset{−\mathrm{1}/\mathrm{2}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\:\mid\:{x}\:\mathrm{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\mid\:{dx}\:= \\ $$ Commented by mathmax by abdo last updated on 21/Feb/20 $${let}\:{I}\:=\int_{−\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 16707 by myaoo5678 last updated on 25/Jun/17 $$\mathrm{If}\:\:{x},\:{y},\:{z}\:\mathrm{are}\:{p}\mathrm{th},\:{q}\mathrm{th}\:\mathrm{and}\:{r}\mathrm{th}\:\mathrm{terms}\:\mathrm{respectively}, \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{and}\:\mathrm{also}\:\mathrm{of}\:\mathrm{GP},\:\mathrm{then}\: \\ $$$${x}^{{y}−{z}} \:{y}^{{z}−{x}} \:{z}^{{x}−{y}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Answered by Tinkutara last updated on…
Question Number 16485 by 09798887650 last updated on 23/Jun/17 $$\mathrm{If}\:\mathrm{tan}^{\mathrm{2}} \theta=\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \phi+\mathrm{1},\:\mathrm{then}\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{sin}^{\mathrm{2}} \phi \\ $$$$\mathrm{equals} \\ $$ Answered by Tinkutara last updated on 23/Jun/17 $$\mathrm{cos}\:\mathrm{2}\theta\:=\:\frac{\mathrm{1}\:−\:\mathrm{tan}^{\mathrm{2}}…