Question Number 170349 by antoineop last updated on 21/May/22 $$ \\ $$$${help}\:{please}\:\: \\ $$$$\overset{\infty} {\int}_{\mathrm{1}} \left(\frac{\mathrm{1}}{{t}}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}}\right)\right)\:{dt} \\ $$$${I}\:{can}\:{show}\:{that}\:{it}\:{is}\:{equal}\:{to} \\ $$$$−\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)}…
Question Number 39280 by rahul 19 last updated on 04/Jul/18 $$\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}\:^{\mathrm{10}} {x}.\mathrm{sin}\:\mathrm{12}{x}\:{dx}\:=\:? \\ $$$$\mathrm{Given} \\ $$$$\mathrm{reduction}\:\mathrm{formula}\:: \\ $$$$\int\:\mathrm{cos}^{\mathrm{m}} {x}\:{sin}\:{nx}\:{dx}\: \\ $$$$\:\mathrm{I}_{\mathrm{m},\mathrm{n}} =\:−\frac{\mathrm{cos}\:^{\mathrm{m}} {x}\:.\:{cosnx}}{{m}+{n}}\:+\:\frac{{m}}{{m}+{n}}\:\mathrm{I}_{\mathrm{m}−\mathrm{1},\mathrm{n}−\mathrm{1}}…
Question Number 39272 by ajfour last updated on 04/Jul/18 $$\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} \frac{\left({a}\mathrm{cos}\:\alpha−{b}\mathrm{cos}\:\theta\right){d}\theta}{\left[{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\left(\theta−\alpha\right)\right]^{\mathrm{3}/\mathrm{2}} }\:=\:? \\ $$ Commented by ajfour last updated on 06/Jul/18 $${please}\:{consider}..…
Question Number 104774 by mathmax by abdo last updated on 23/Jul/20 $$\mathrm{let}\:\mathrm{B}_{\mathrm{n}} =\:\int\int_{\left[\mathrm{0},\mathrm{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} }}\mathrm{dxdy} \\ $$$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\mathrm{B}_{\mathrm{n}} }{\mathrm{n}} \\ $$…
Question Number 104773 by mathmax by abdo last updated on 23/Jul/20 $$\mathrm{let}\:\mathrm{A}_{\mathrm{n}} =\int\int_{\left[\mathrm{0},\mathrm{n}\left[^{\mathrm{2}} \right.\right.} \:\:\:\:\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} } }{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} }}\mathrm{dxdy} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculste}\:\mathrm{A}_{\mathrm{n}} \:\mathrm{interm}\:\mathrm{of}\:\mathrm{n} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty}…
Question Number 39230 by MJS last updated on 04/Jul/18 $$\int\frac{{d}\alpha}{\mathrm{sin}\:\mathrm{2}\alpha\:+\mathrm{sin}\:\mathrm{3}\alpha\:+\mathrm{sin}\:\mathrm{5}\alpha}=? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jul/18 $${sin}\mathrm{2}\alpha+{sin}\mathrm{3}\alpha+{sin}\mathrm{5}\alpha \\ $$$${sin}\mathrm{2}\alpha+{sin}\mathrm{3}\alpha+{sin}\mathrm{2}\alpha{cos}\mathrm{3}\alpha+{cos}\mathrm{2}\alpha{sin}\mathrm{3}\alpha \\ $$$${sin}\mathrm{2}\alpha\left(\mathrm{1}+{cos}\mathrm{3}\alpha\right)+{sin}\mathrm{3}\alpha\left(\mathrm{1}+{cos}\mathrm{2}\alpha\right) \\…
Question Number 39214 by math khazana by abdo last updated on 03/Jul/18 $${let}\:{f}\left({x}\right)=\:\frac{{e}^{−\mathrm{3}{x}} }{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${developp}\:{f}\:\:{at}\:{integr}\:{serie}. \\ $$ Commented by math khazana by abdo…
Question Number 104732 by Dwaipayan Shikari last updated on 23/Jul/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({tan}\theta\right){d}\theta \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms of…
Question Number 170255 by mathlove last updated on 19/May/22 $$\frac{{d}}{{dx}}\left[\int_{\mathrm{2}} ^{{x}} {e}^{{t}^{\mathrm{2}} } {dt}=?\right. \\ $$ Answered by floor(10²Eta[1]) last updated on 19/May/22 $$\int_{\mathrm{2}} ^{\mathrm{x}}…
Question Number 104718 by M±th+et+s last updated on 23/Jul/20 $$\int{tan}^{−\mathrm{1}} \left(\frac{{a}\sqrt{{x}}+{b}}{{c}}\right){dx}\: \\ $$ Answered by Dwaipayan Shikari last updated on 23/Jul/20 $$\int\frac{\mathrm{2}{c}\sqrt{{x}}}{{a}}{tan}^{−\mathrm{1}} {udu}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}=\frac{{a}\sqrt{{x}}+{b}}{{c}}\Rightarrow\frac{{a}}{\mathrm{2}{c}\sqrt{{x}}}=\frac{{du}}{{dx}} \\ $$$$\frac{\mathrm{2}{c}}{{a}}\int\sqrt{{x}}{tan}^{−\mathrm{1}}…