Question Number 190168 by TUN last updated on 29/Mar/23 $$\left.\mathrm{1}\right)\underset{\mathrm{0}} {\int}^{\infty} \frac{{sin}\:{x}}{{x}^{{p}} +{sin}\:{x}}{dx}\:,{p}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\underset{\pi} {\int}^{\infty} \frac{{xcos}\:{x}}{{x}^{{p}} +{x}^{{q}} }{dx},{p}>\mathrm{0}{and}\:{q}>\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\underset{\mathrm{0}} {\int}^{\infty} \frac{{sin}\:{x}^{{p}} }{\:{x}^{{q}} }{dx},\:{p}>\mathrm{0},{q}>\mathrm{0}…
Question Number 124632 by liberty last updated on 05/Dec/20 $${Calculate}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\:{dx}\: \\ $$ Answered by bemath last updated on 05/Dec/20 $${put}\:{x}=\mathrm{2cos}\:\mathrm{2}\theta\: \\ $$$${I}=\underset{\pi/\mathrm{4}} {\overset{\mathrm{0}}…
Question Number 124624 by bemath last updated on 04/Dec/20 $$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{x}}\:+\mathrm{4}{x}} \\ $$ Answered by Dwaipayan Shikari last updated on 04/Dec/20 $$\int\frac{\mathrm{3}{u}^{\mathrm{2}} {du}}{{u}+\mathrm{4}{u}^{\mathrm{3}} }=\frac{\mathrm{3}}{\mathrm{8}}\int\frac{\mathrm{8}{udu}}{\mathrm{1}+\mathrm{4}{u}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{8}}{log}\left(\mathrm{1}+\mathrm{4}{u}^{\mathrm{2}} \right)=\frac{\mathrm{3}}{\mathrm{8}}{log}\left(\mathrm{1}+\mathrm{4}{x}^{\frac{\mathrm{2}}{\mathrm{3}}}…
Question Number 124608 by TANMAY PANACEA last updated on 04/Dec/20 $$\int_{\mathrm{0}} ^{\infty} {sinx}^{{p}} \:{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}^{{p}} }{{x}^{{q}} }{dx} \\ $$$${collected}\:{question} \\ $$ Answered…
Question Number 124607 by Mammadli last updated on 04/Dec/20 $$\int\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by MJS_new last updated on 04/Dec/20 $$\left(\mathrm{1}\right)\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}: \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}}…
Question Number 124594 by physicstutes last updated on 04/Dec/20 $$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\mathrm{ln}\:\mathrm{2}} \frac{\mathrm{1}}{\mathrm{cosh}\left({x}\:+\:\mathrm{ln}\:\mathrm{4}\right)}\:{dx}\:=\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{33}}\right) \\ $$ Commented by mohammad17 last updated on 04/Dec/20 $${let}:{y}={x}+{ln}\mathrm{4}\Rightarrow{dy}={dx}…
Question Number 59052 by maxmathsup by imad last updated on 04/May/19 $${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta\:+{xsin}\theta\right){d}\theta\:\:\:{with}\:{x}\:{fromR} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta\:+{sin}\theta\right)\:{d}\theta\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta\:+\mathrm{2}{sin}\theta\right){d}\theta\:. \\ $$ Terms…
Question Number 59050 by maxmathsup by imad last updated on 04/May/19 $${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\: \\ $$ Answered by tanmay last updated on 04/May/19…
Question Number 124587 by mnjuly1970 last updated on 04/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\blacktriangleleft::::\blacktriangleright\:{calculus} \\ $$$$\:\:\:\:\:{simple}\:\:{question}:: \\ $$$$\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{4}}{\:\sqrt{\mathrm{4}+{x}^{\mathrm{4}} }}\:{dx}\overset{???} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{{sin}\left({x}\right)}}\:+\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{{cos}\left({x}\right)}} \\…
Question Number 124579 by bemath last updated on 04/Dec/20 $$\:\:{I}=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{dx}}{\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }\:? \\ $$ Answered by liberty last updated on 04/Dec/20 $${I}=\int_{\:\mathrm{0}} ^{\:\infty}…