Question Number 43159 by MASANJA J last updated on 07/Sep/18 Answered by alex041103 last updated on 08/Sep/18 $${For}\:\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{15}}}\:: \\ $$$$\int\frac{{dx}}{\:\sqrt{{x}+\mathrm{15}}}\:=\:\int\frac{{d}\left({x}+\mathrm{15}\right)}{\:\sqrt{{x}+\mathrm{15}}}=\int{u}^{−\mathrm{1}/\mathrm{2}} {du}= \\ $$$$=\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}} +{C}=\mathrm{2}\sqrt{{x}+\mathrm{15}}\:+{C} \\…
Question Number 108692 by 150505R last updated on 18/Aug/20 Answered by Dwaipayan Shikari last updated on 18/Aug/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{log}\left({tan}\theta\right)}{{tan}^{\mathrm{2}} \theta+\mathrm{1}}{sec}^{\mathrm{2}} \theta{d}\theta\:\:\:\left({x}={tan}\theta,\:\mathrm{1}={sec}^{\mathrm{2}} \theta\frac{{d}\theta}{{dx}}\right) \\ $$$$\mathrm{2}\int_{\mathrm{0}}…
Question Number 43156 by MASANJA J last updated on 07/Sep/18 $${integrate}\:{w}.{r}.{t}\:{x} \\ $$$$\int\frac{{xe}^{{x}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18 $${x}={tan}\alpha\:\:{dx}={sec}^{\mathrm{2}}…
Question Number 43157 by MASANJA J last updated on 07/Sep/18 $$\int{secxdx} \\ $$ Commented by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{J}\:=\:\int\:\:\frac{{dx}}{{cosx}}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}\:{give} \\ $$$${J}\:=\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}}…
Question Number 43147 by rahul 19 last updated on 07/Sep/18 $$\mathrm{If}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}\:}\:, \\ $$$$\mathrm{then}\:{prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{a}{x}^{\mathrm{2}} } {dx}\:=\:\sqrt{\frac{\pi}{\mathrm{4a}}} \\ $$$$\mathrm{where}\:\mathrm{a}>\mathrm{0}. \\ $$…
Question Number 43145 by rahul 19 last updated on 07/Sep/18 $$\int_{\mathrm{0}} ^{\infty} \:\left[\:\mathrm{2e}^{−{x}} \right]{dx}\:=\:?\: \\ $$$${where}\:\left[.\right]=\:{gif}. \\ $$ Commented by maxmathsup by imad last updated…
Question Number 108667 by bobhans last updated on 18/Aug/20 Answered by john santu last updated on 18/Aug/20 Commented by john santu last updated on 18/Aug/20…
Question Number 108663 by bobhans last updated on 18/Aug/20 Answered by john santu last updated on 18/Aug/20 $$\:\:\:\:\:\frac{\frac{\multimap{J}}{{S}\leftrightharpoons}}{} \\ $$$${let}\:\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }\:=\:{m}\:\Rightarrow{x}^{\mathrm{3}} \:=\:{m}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}}…
Question Number 43125 by Raj Singh last updated on 07/Sep/18 Commented by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{I}\:=\int\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{{sinx}}+\frac{\mathrm{1}}{{cosx}}}{dx}\:\Rightarrow{I}\:=\:\int\:\:\:\:\frac{{cosx}\:{sinx}}{{cosx}\:+{sinx}}{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{2}}{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{dx}\:\:{changement}\:\:{x}−\frac{\pi}{\mathrm{4}}={t}\:{give}\: \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\frac{{sin}\left(\mathrm{2}\left({t}+\frac{\pi}{\mathrm{4}}\right)\right)}{{cost}}\:{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{{cost}}\:{dt} \\…
Question Number 43100 by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cos}\theta}{\mathrm{1}+{xsin}\theta}{d}\theta \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}\left(\mathrm{2}\theta\right)}{\left(\mathrm{1}+{xsin}\theta\right)^{\mathrm{2}} }{d}\theta \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{cos}\theta}{\mathrm{1}+\mathrm{2}{cos}\theta}{d}\theta\:\:\:{and}\:\:\int_{\mathrm{0}}…