Question Number 94609 by M±th+et+s last updated on 20/May/20 $$\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{\mathrm{2}{x}+\mathrm{3}}}{dx} \\ $$ Answered by mathmax by abdo last updated on 20/May/20 $$\mathrm{I}\:=\int\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}+\sqrt{\mathrm{2x}+\mathrm{3}}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int\:\frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\sqrt{\mathrm{2x}+\mathrm{3}}−\sqrt{\mathrm{x}+\mathrm{1}}\right)}{\mathrm{x}+\mathrm{2}}\mathrm{dx}…
Question Number 29043 by yesaditya22@gmail.com last updated on 03/Feb/18 $$\int\mathrm{tan}^{−} \left(\mathrm{1}−\mathrm{sinx}/\mathrm{1}+\mathrm{sinx}\right)\:\mathrm{dx} \\ $$ Commented by abdo imad last updated on 03/Feb/18 $${let}\:{put}\:{I}=\:\int\:{arctan}\left(\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}\right){dx}\:\:\:\left({arctan}={tan}^{−\mathrm{1}} \right){we}\:{have} \\ $$$$\frac{\mathrm{1}−{sinx}}{\mathrm{1}+{sinx}}=\frac{\mathrm{1}\:−{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}=\frac{\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 160113 by mnjuly1970 last updated on 25/Nov/21 $$ \\ $$$$\:\:{prove}\:\: \\ $$$$\:\:\:\:\:\:\Phi:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sinh}\left({x}\right)}{{cosh}^{\mathrm{2}} \left({x}\right)}.\frac{\mathrm{1}}{{x}}\:{dx}\:\overset{???} {=}\:\frac{\mathrm{4G}}{\pi} \\ $$ Answered by mindispower last updated…
Question Number 29038 by abdo imad last updated on 03/Feb/18 $${find}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left({at}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}. \\ $$ Commented by abdo imad last updated on 11/Feb/18 $${let}\:{put}\:{I}\left({a}\right)=\:\underset{−\infty}…
Question Number 94574 by student work last updated on 19/May/20 $$\int\frac{\sqrt{\mathrm{cosx}}}{\:\sqrt{\mathrm{sinx}\:}\:+\sqrt{\mathrm{cosx}}}\mathrm{dx}=? \\ $$ Commented by student work last updated on 19/May/20 $$\mathrm{please}\:\mathrm{solve}\:\mathrm{who}\:\mathrm{can}? \\ $$ Answered…
Question Number 29027 by abdo imad last updated on 03/Feb/18 $${find}\:\int\int_{{D}} \:{e}^{−{y}} {sin}\left(\mathrm{2}{xy}\right){dxdy}\:{with}\:{D}=\left[\mathrm{0},\mathrm{1}\right]×\left[\mathrm{0},+\infty\left[\right.\right. \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{2}} {t}}{{t}}\:{e}^{−{t}} {dt}\:\:. \\ $$ Terms of Service Privacy…
Question Number 29028 by abdo imad last updated on 03/Feb/18 $${for}\:{t}>\mathrm{0}\:\:{and}\:{f}\left({t}\right)=\:\left(\mathrm{4}\pi{t}\right)^{−\frac{{n}}{\mathrm{2}}} \:\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}{t}}} \:\:\:{prove}\:{that} \\ $$$$\int_{{R}} {f}_{{t}} \left({x}\right){dx}=\mathrm{1}\:\:\:\forall{t}>\mathrm{0}. \\ $$ Terms of Service Privacy Policy…
Question Number 29018 by Joel578 last updated on 03/Feb/18 $$\int\:\sqrt{\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{n}} \:\mathrm{tan}^{\mathrm{2}{n}} \:\left(\mathrm{2}{x}\right)\right]}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94544 by Ar Brandon last updated on 19/May/20 $$\mathrm{1}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]\:\mathrm{is}\:\mathrm{of}\:\mathrm{Riemann} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{segments}\:\mathrm{of}\:\mathbb{R} \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{defined}\:\mathrm{within}\:\mathrm{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{1}\:\mathrm{if}\:\mathrm{x}\in\mathbb{Q}\cap\left[\mathrm{0},\mathrm{1}\right]}\\{\mathrm{0}\:\:\mathrm{otherwise}}\end{cases}\:\:\mathrm{is}\:\mathrm{not}\:\mathrm{of}\:\mathrm{Riemann}\:\mathrm{on}\:\mathrm{x}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 29002 by abdo imad last updated on 03/Feb/18 $${let}\:{give}\:\mathrm{0}<{p}<\mathrm{1}\:{calculate}\:\:{K}\left({p}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{pt}} }{\mathrm{1}+{e}^{{t}} }{dt}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com