Question Number 92421 by mhmd last updated on 06/May/20 $$\int\frac{{cscx}}{{cos}\left(\mathrm{2}{x}\right)+\mathrm{2}{cos}^{\mathrm{2}} {x}}{dx} \\ $$$${pleas}\:{sir}\:{help}\:{me} \\ $$ Answered by niroj last updated on 07/May/20 $$\:\:\int\:\:\frac{\:\:\mathrm{cosec}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{2x}+\:\mathrm{2cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\: \\…
Question Number 92410 by mathmax by abdo last updated on 06/May/20 $${find}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+\mathrm{3}{x}}−\sqrt{\mathrm{1}−\mathrm{3}{x}}} \\ $$ Commented by Prithwish Sen 1 last updated on 06/May/20…
Question Number 92407 by mathmax by abdo last updated on 06/May/20 $${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{\mathrm{1}+{x}}+{a}\sqrt{\mathrm{1}−{x}}\right){dx}\:\:\:{with}\:\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){explicite}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+{a}\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{\mathrm{1}+{x}}+\mathrm{2}\sqrt{\mathrm{1}−{x}}\right){dx} \\…
Question Number 157932 by cortano last updated on 30/Oct/21 $${prove}\:{that}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xy}}{{rz}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{xz}}{{ry}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{yz}}{{rx}}\right)=\frac{\pi}{\mathrm{2}} \\ $$ Commented by cortano last updated on 30/Oct/21 $${no}\:{sir}.\:{only}\:{it}\:{condition} \\ $$…
Question Number 157929 by akolade last updated on 30/Oct/21 $$\int\frac{\mathrm{dx}}{\mathrm{sin}\:\mathrm{x}+\:\mathrm{sec}\:\mathrm{x}} \\ $$$$\mathrm{using}\:\mathrm{wiestress}\:\mathrm{substitution} \\ $$ Commented by cortano last updated on 30/Oct/21 $${C}\:=\:\int\:\frac{\mathrm{1}}{\:\mathrm{sin}\:{x}+\mathrm{sec}\:{x}}\:{dx}\: \\ $$$$\:\left[\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\:{u}\:\rightarrow\begin{cases}{\mathrm{sin}\:{x}\:=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\\{\mathrm{cos}\:{x}=\frac{\mathrm{1}−{u}^{\mathrm{2}}…
Question Number 92397 by john santu last updated on 06/May/20 $$\int\:\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\: \\ $$$$\left[\:{x}\:=\:\mathrm{sin}\:{w}\:\right]\: \\ $$$$\int\:\frac{\mathrm{cos}\:\mathrm{w}\:\mathrm{dw}}{\mathrm{sin}\:\mathrm{w}−\mathrm{cos}\:\mathrm{w}}\:=\:\int\:\frac{\mathrm{dw}}{\mathrm{tan}\:\mathrm{w}−\mathrm{1}} \\ $$$$=\:\int\:\frac{\mathrm{sec}^{\mathrm{2}} \:\mathrm{w}\:\mathrm{dw}}{\left(\mathrm{tan}\:\mathrm{w}−\mathrm{1}\right)\mathrm{sec}^{\mathrm{2}} \:\mathrm{w}} \\ $$$$=\:\int\:\frac{\mathrm{du}}{\left(\mathrm{u}−\mathrm{1}\right)\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}\:;\:\left[\:\mathrm{u}\:=\:\mathrm{tan}\:\mathrm{w}\:\right]\: \\ $$$$=\:\int\:\frac{\mathrm{du}}{\mathrm{2}\left(\mathrm{u}−\mathrm{1}\right)}−\int\:\frac{\mathrm{u}\:\mathrm{du}\:}{\mathrm{2}\left(\mathrm{u}^{\mathrm{2}}…
Question Number 92394 by john santu last updated on 06/May/20 $$\int\:\mathrm{ln}\:\left(\sqrt{\mathrm{1}−{x}}\:+\:\sqrt{\mathrm{1}+{x}}\:\right)\:{dx}\: \\ $$ Commented by mathmax by abdo last updated on 06/May/20 $${I}\:=\int{ln}\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{1}+{x}}\right){dx}\:\:{by}\:{parts}\: \\ $$$${I}\:={x}\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{1}+{x}}\right)−\int\:{x}\frac{\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{1}+{x}}\right)^{'}…
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Question Number 157870 by Odhiambojr last updated on 29/Oct/21 $${find}\:{the}\:{integral}: \\ $$$$\int\left\{\left(\mathrm{3}{x}+\mathrm{1}\right)/\left({x}^{\mathrm{2}} +\mathrm{4}\right)\right\}{dx} \\ $$ Answered by puissant last updated on 29/Oct/21 $$\Omega=\int\:\frac{\mathrm{3}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx}\:=\:\int\frac{\mathrm{3}{x}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx}+\int\frac{{dx}}{{x}^{\mathrm{2}}…