Question Number 84496 by bshahid010@gmail.com last updated on 13/Mar/20 Commented by mathmax by abdo last updated on 13/Mar/20 $${I}\:=\int\:\:\frac{{arcsinx}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts}\:\:{u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and}\:{v}={arcsinx}\:\Rightarrow \\ $$$${I}\:=−\frac{{arsinx}}{{x}}\:−\int−\frac{\mathrm{1}}{{x}}\:×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}…
Question Number 149996 by tabata last updated on 08/Aug/21 $$\left(\mathrm{1}\right)\:\int\:\:\frac{{dx}}{\mathrm{1}+{tanx}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\int\:\:\frac{\sqrt{{tanx}}}{{sinx}\:{cosx}}{dx} \\ $$ Answered by mindispower last updated on 08/Aug/21 $$=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}.\frac{\sqrt{{tg}\left({x}\right)}}{{tg}\left({x}\right)}{dx}=\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}}…
Question Number 149989 by ZiYangLee last updated on 08/Aug/21 $$\mathrm{By}\:\mathrm{subs}\:{u}^{\mathrm{2}} =\mathrm{4}+{x},\:\mathrm{evaluate}\:\int\:\frac{\sqrt{\mathrm{4}+{x}}}{{x}}\:{dx} \\ $$ Answered by puissant last updated on 08/Aug/21 $$\int\frac{\sqrt{\mathrm{4}+{x}}}{{x}}{dx}={Q} \\ $$$${u}=\sqrt{\mathrm{4}+{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{4}+{x}\:\rightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{4}…
Question Number 149959 by Mathfinity last updated on 08/Aug/21 Commented by john_santu last updated on 08/Aug/21 Commented by john_santu last updated on 08/Aug/21 Commented by…
Question Number 149946 by DELETED last updated on 08/Aug/21 Answered by DELETED last updated on 08/Aug/21 $$\boldsymbol{\mathrm{Jawaban}}\:\boldsymbol{\mathrm{no}}\:\mathrm{2} \\ $$$$\mathrm{Gunakan}\:\mathrm{Hk}\:\mathrm{Kirchoff}\:\mathrm{I} \\ $$$$\Sigma\mathrm{I}_{\mathrm{masuk}} =\Sigma\mathrm{I}_{\mathrm{keluar}} \\ $$$$\mathrm{I}_{\mathrm{1}} +\mathrm{I}_{\mathrm{2}}…
Question Number 18872 by mondodotto@gmail.com last updated on 31/Jul/17 Answered by ajfour last updated on 31/Jul/17 $$\mathrm{b}=\mathrm{a}+\frac{\mathrm{1}}{\mathrm{b}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{b}−\frac{\mathrm{1}}{\mathrm{b}}\: \\ $$ Answered by Joel577 last…
Question Number 18866 by mondodotto@gmail.com last updated on 31/Jul/17 Answered by dioph last updated on 31/Jul/17 $$\mathrm{Define}\:{y}\:\equiv\:\mathrm{log}\:{x} \\ $$$$\mathrm{Substitute}\:{x}\:=\:\mathrm{10}^{{y}} \:\left(\mathrm{here}\:\mathrm{I}\:\mathrm{assumed}\right. \\ $$$$\left.\mathrm{base}\:\mathrm{10}\:\mathrm{but}\:\mathrm{the}\:\mathrm{idea}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{for}\:\mathrm{any}\:\mathrm{base}\right) \\ $$$$\mathrm{5}^{{y}} \:+\:\left(\mathrm{10}^{{y}}…
Question Number 149914 by 7770 last updated on 08/Aug/21 $$\boldsymbol{{show}}\:\boldsymbol{{that}}\int_{\mathrm{1}} ^{\mathrm{2}} \left(\frac{\mathrm{2}+\mathrm{6}\theta^{\mathrm{2}} −\mathrm{2}\boldsymbol{\theta}^{\mathrm{3}} }{\boldsymbol{\theta}^{\mathrm{2}} \left(\boldsymbol{\theta}^{\mathrm{2}} +\mathrm{1}\right)}\right)\boldsymbol{{d}\theta}=\mathrm{1}.\mathrm{606} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18834 by mondodotto@gmail.com last updated on 30/Jul/17 Answered by Tinkutara last updated on 30/Jul/17 $$\mathrm{tan}\:\mathrm{2A}\:=\:\mathrm{cot}\:\left(\mathrm{90}°\:−\:\mathrm{2A}\right)\:=\:\mathrm{cot}\:\left(\mathrm{A}\:−\:\mathrm{18}°\right) \\ $$$$\mathrm{Since}\:\mathrm{A}\:\mathrm{is}\:\mathrm{acute},\:\mathrm{so}\:\mathrm{cot}\:\mathrm{A}\:=\:\mathrm{cot}\:\mathrm{B}\:\Leftrightarrow\:\mathrm{A}\:=\:\mathrm{B} \\ $$$$\mathrm{90}°\:−\:\mathrm{2A}\:=\:\mathrm{A}\:−\:\mathrm{18}° \\ $$$$\mathrm{A}\:=\:\mathrm{36}° \\ $$…
Question Number 18827 by aplus last updated on 30/Jul/17 Commented by aplus last updated on 30/Jul/17 $$\mathrm{help}\:\mathrm{guys} \\ $$ Commented by mrW1 last updated on…