Question Number 40505 by prof Abdo imad last updated on 23/Jul/18 $${calcilate}\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)\:+{cos}\left(\mathrm{2}{x}\right)}{dx} \\ $$ Answered by MJS last updated on 23/Jul/18 $$\mathrm{cos}\:\mathrm{2}{x}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:−\mathrm{1}…
Question Number 171565 by Tawa11 last updated on 17/Jun/22 Commented by infinityaction last updated on 17/Jun/22 $$\:\:\:\:{p}\:\:=\:\:\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\:\:\:{then}\:\:\:\frac{\mathrm{1}}{{p}}\:\:=\:\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\left({p}^{\mathrm{2}} \:+\:\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\:\:−\mathrm{2}\right)^{\mathrm{1}/\mathrm{2}} =\:\:\left({p}−\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\int_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}…
Question Number 171560 by mnjuly1970 last updated on 17/Jun/22 $$ \\ $$$$\:\:\:\:\:\mathrm{Nice}\:\:\:\mathrm{Integral} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\:{tan}\left({x}\right)}{\left(\:{cos}^{\:\mathrm{2}} \left({x}\right)\:\:+\:\mathrm{2}{sin}^{\:\mathrm{2}} \left({x}\right)\right)}{dx}\:= \\ $$ Commented by infinityaction…
Question Number 106010 by bemath last updated on 02/Aug/20 $$\int\:\frac{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}\:{dx}\: \\ $$ Answered by Dwaipayan Shikari last updated on 02/Aug/20 $$\int\frac{\mathrm{1}+{x}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}+\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}}…
Question Number 40467 by gunawan last updated on 22/Jul/18 $$\int\mathrm{ln}\:\mid\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}\mid\:{dx}= \\ $$ Commented by maxmathsup by imad last updated on 22/Jul/18 $${let}\:{I}\:=\:\int\:{ln}\mid\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}}\mid{dx}\:\:{due}\:{to}\:{x}\geqslant\mathrm{0}\:{we}\:{have} \\ $$$${I}\:=\:\int\:{ln}\left(\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}\right){dx}\:=_{\sqrt{{x}}={t}} \:\mathrm{2}\:\:\int{ln}\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 106001 by Algoritm last updated on 02/Aug/20 Answered by JDamian last updated on 02/Aug/20 $$\mathrm{5056}\left({e}−\mathrm{1}\right) \\ $$ Answered by mr W last updated…
Question Number 105983 by bemath last updated on 02/Aug/20 $$\int\:\frac{\mathrm{sin}\:{x}\:{dx}}{\mathrm{6}−\mathrm{sin}\:^{\mathrm{2}} {x}}\:?\:\: \\ $$ Commented by bemath last updated on 02/Aug/20 $$\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{6}−\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)}\:{dx}\:=\:−\int\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\mathrm{5}+\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$${let}\:\flat\:=\:\mathrm{cos}\:{x}…
Question Number 40442 by rahul 19 last updated on 21/Jul/18 $$\mathrm{Solve}\:: \\ $$$$\mathrm{y}^{\mathrm{4}} \mathrm{d}{x}\:+\:\mathrm{2}{x}\mathrm{y}^{\mathrm{3}} \mathrm{dy}\:=\:\frac{\mathrm{yd}{x}−\:{x}\mathrm{dy}}{{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} }. \\ $$ Answered by ajfour last updated on…
Question Number 105969 by Ar Brandon last updated on 02/Aug/20 $$\int\frac{\mathrm{sinx}+\mathrm{cosx}}{\:\sqrt{\mathrm{1}+\mathrm{sin2x}}}\mathrm{dx} \\ $$ Answered by bobhans last updated on 02/Aug/20 $$\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }}\:\mathrm{dx}\:=\:\int\frac{\:\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\mathrm{dx}}{\mid\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\mid} \\ $$$$=\:\pm\:\int\:\mathrm{dx}\:=\:\pm\mathrm{x}\:+\:\mathrm{C} \\…
Question Number 40422 by rahul 19 last updated on 21/Jul/18 $$\mathrm{Solve}: \\ $$$$\mathrm{yd}{x}\:−\:{xdy}\:+\mathrm{log}\:{xdx}\:=\mathrm{0} \\ $$ Commented by prof Abdo imad last updated on 22/Jul/18 $$\Rightarrow{y}\:\:−{x}\frac{{dy}}{{dx}}\:+{ln}\left({x}\right)=\mathrm{0}\:\Rightarrow{xy}^{'}…