Question Number 161285 by cortano last updated on 15/Dec/21 $$\left(\mathrm{1}\right)\:\int\:\frac{{dx}}{\mathrm{1}−\mathrm{2cos}\:{x}} \\ $$$$\left(\mathrm{2}\right)\:\int\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:{dx} \\ $$$$\left(\mathrm{3}\right)\:\int\:\frac{{dx}}{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{sin}\:{x}} \\ $$ Answered by bobhans last updated on 15/Dec/21 $$\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{dx}}{\mathrm{1}−\mathrm{2cos}\:\mathrm{x}}\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{1}−\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}}…
Question Number 30216 by abdo imad last updated on 18/Feb/18 $${let}\:{I}\left({x}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{{x}^{\mathrm{2}} \:+{cos}^{\mathrm{2}} {t}} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{I}\left({x}\right)=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{{x}^{\mathrm{2}} \:+{cos}^{\mathrm{2}} {t}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{I}\left({x}\right). \\ $$…
Question Number 30215 by abdo imad last updated on 18/Feb/18 $${let}\:{give}\:{J}\left({x}\right)=\:\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\pi} {cos}\left({xcost}\right){dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{J}^{'} \:{and}\:{J}^{''} \:{in}\:{form}\:{of}\:{integrals} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:{J}^{'} \left({x}\right)=\frac{−{x}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{sin}^{\mathrm{2}} {t}\:{cos}\left({xcost}\right){dt}\:{and}\:{J}\:{is} \\ $$$${solution}\:{of}\:{d}.{e}.\:\:{xy}^{''}…
Question Number 161281 by mnjuly1970 last updated on 15/Dec/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 95738 by bobhans last updated on 27/May/20 $$\int\:\frac{\mathrm{p}−\mathrm{tan}\:\mathrm{x}}{\mathrm{p}+\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$ Commented by PRITHWISH SEN 2 last updated on 27/May/20 $$\frac{\mathrm{pcosx}−\mathrm{sin}\:\mathrm{x}}{\mathrm{pcos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}\:=\:\frac{\mathrm{m}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{psin}\:\mathrm{x}\right)+\mathrm{n}\left(\mathrm{pcos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{pcos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} \\ $$$$\mathrm{m}=\frac{\mathrm{2p}}{\mathrm{1}+\mathrm{p}^{\mathrm{2}} }…
Question Number 161265 by mathlove last updated on 15/Dec/21 Commented by amin96 last updated on 15/Dec/21 $${OMG}\left(\:\:\:{e}^{{x}} ={u}\:\:{e}^{{u}} ={t}\:\:\:{e}^{{t}} ={v}………….\right. \\ $$ Commented by MJS_new…
Question Number 95722 by rb222 last updated on 27/May/20 $${use}\:{cylinder}\:{ring}\:{method} \\ $$$$ \\ $$$${y}\:=\:\mathrm{2}{x}−\mathrm{1} \\ $$$${y}\:=\:−\mathrm{2}{x}\:+\:\mathrm{3} \\ $$$${x}\:=\:\mathrm{2}\: \\ $$$$ \\ $$$${y}−{axis}\: \\ $$$$ \\…
Question Number 161256 by cortano last updated on 15/Dec/21 $$\:{Given}\:{f}\left({x}\right)={f}\left({x}+\mathrm{2}\right),\:\forall{x}\in\mathbb{R} \\ $$$$\:{If}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}=\:{p}\:{then}\:\underset{\mathrm{0}} {\overset{\mathrm{2020}} {\int}}{f}\left({x}+\mathrm{2}{a}\right){dx}=? \\ $$$$\:{for}\:{a}\in\mathbb{Z}^{+} \\ $$ Answered by talminator2856791 last updated…
Question Number 30185 by abdo imad last updated on 17/Feb/18 $${let}\:\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{sinx}}{\:\sqrt{\mathrm{1}+{sinxcosx}}}{dx}\:{and} \\ $$$${J}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{cosx}}{\:\sqrt{\mathrm{1}+{sinx}\:{cosx}}}\:{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}\:+{J} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{I}\:{and}\:{J}. \\ $$ Terms of…
Question Number 30182 by abdo imad last updated on 17/Feb/18 $${find}\:\int_{\mathrm{2}} ^{\mathrm{3}} \:\:\:\frac{\sqrt{{x}+\mathrm{1}}}{{x}\sqrt{\mathrm{1}−{x}}}{dx}\:. \\ $$ Commented by abdo imad last updated on 21/Feb/18 $${let}\:{put}\:\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:={t}\:\Leftrightarrow\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\:={t}^{\mathrm{2}} \:\:\Leftrightarrow{x}+\mathrm{1}=−{t}^{\mathrm{2}}…