Question Number 54529 by tarun kunar last updated on 05/Feb/19 $$\int\mathrm{5cos}\:{x}−\mathrm{4sin}\:{x}/\mathrm{2cos}\:{x}+\mathrm{sin}\:{x}\:{dx} \\ $$ Commented by maxmathsup by imad last updated on 05/Feb/19 $${let}\:{I}\:=\int\:\frac{\mathrm{5}{cosx}−\mathrm{4}{sinx}}{\mathrm{2}{cosx}\:+{sinx}}{dx}\:\:{let}\:{use}\:{the}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{\frac{\mathrm{5}\left(\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 120060 by bemath last updated on 29/Oct/20 $$\left({i}\right)\:\underset{−\mathrm{2}} {\overset{\mathrm{0}} {\int}}\:\frac{{dx}}{\mathrm{2}{x}+\mathrm{3}} \\ $$$$\left({ii}\right)\underset{\mathrm{3}} {\overset{\mathrm{5}} {\int}}\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{4}−{x}\right)^{\mathrm{2}} }}\: \\ $$ Answered by bramlexs22 last updated on…
Question Number 120059 by A8;15: last updated on 29/Oct/20 Answered by Lordose last updated on 29/Oct/20 $$ \\ $$$$\mathrm{Firstly},\:\mathrm{compute}\:\mathrm{the}\:\mathrm{indefinite}\:\mathrm{integral} \\ $$$$\mathrm{I}\:=\:\int\mathrm{xsec}\left(\mathrm{2x}\right)\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{asec}\left(\mathrm{a}\right)\mathrm{da}\:\left\{\mathrm{a}=\mathrm{2x}\right\} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{alntan}\left(\frac{\mathrm{a}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\:−\:\int\mathrm{lntan}\left(\frac{\mathrm{a}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mathrm{da}\right)\:\:\:\left\{\mathrm{IBP}\right\} \\ $$$$\mathrm{set}\:\frac{\mathrm{a}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\:=\:\mathrm{y}\:\Rightarrow\:\mathrm{da}=\mathrm{2dy}…
Question Number 120040 by bramlexs22 last updated on 28/Oct/20 $$\:\int\:\frac{{t}^{\mathrm{5}} }{\:\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }}\:{dt}\: \\ $$ Answered by john santu last updated on 28/Oct/20 $$\:{let}\:\mathrm{2}+{t}^{\mathrm{2}} \:=\:{h}^{\mathrm{2}} \:\rightarrow\:{t}\:{dt}\:=\:{h}\:{dh}\:…
Question Number 120025 by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{calculate}\:\mathrm{f}^{'} \left(\mathrm{x}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}}…
Question Number 119989 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xdx}}{\mathrm{2e}^{\mathrm{x}} −\mathrm{1}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xe}^{−\mathrm{x}} \mathrm{dx}}{\mathrm{2}−\mathrm{e}^{−\mathrm{x}}…
Question Number 119970 by bramlexs22 last updated on 28/Oct/20 $$\:\int\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\:? \\ $$ Answered by bemath last updated on 28/Oct/20 $$\:\int\:\frac{{dx}}{{x}^{\mathrm{3}} \sqrt{\frac{\mathrm{25}}{{x}^{\mathrm{2}} }−\mathrm{1}}}\:{dx}\:=\:\int\:\frac{{x}^{−\mathrm{3}} }{\:\sqrt{\mathrm{25}{x}^{−\mathrm{2}}…
Question Number 119960 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\left(\mathrm{1}+\mathrm{x}\right)\sqrt{\mathrm{x}}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}}…
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Question Number 185484 by Mingma last updated on 22/Jan/23 Answered by Ar Brandon last updated on 22/Jan/23 $${I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{\mathrm{3}{x}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} {e}^{−\mathrm{3}{x}}…