Question Number 91326 by mhmd last updated on 29/Apr/20 Commented by mathmax by abdo last updated on 30/Apr/20 $${I}\:=\int\:\:\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:\Rightarrow\:{I}\:=_{{x}−\mathrm{1}={t}} \:\:\:\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} \sqrt{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\left({t}+\mathrm{1}\right)+\mathrm{1}}} \\…
Question Number 156860 by amin96 last updated on 16/Oct/21 $$\int_{\mathrm{0}} ^{\infty} \frac{\sqrt[{{n}}]{{x}}}{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=? \\ $$ Answered by mindispower last updated on 16/Oct/21 $${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty}…
Question Number 156849 by amin96 last updated on 16/Oct/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({e}+\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)}{\:\sqrt{{t}}}{dt}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 91310 by M±th+et+s last updated on 29/Apr/20 $${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cot}\:\left({x}\right)\:{ln}\left({sec}\left({x}\right)\right)\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 25777 by abhishekkumar22121999@gmail.co last updated on 14/Dec/17 $$\int\frac{\mathrm{1}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{dx} \\ $$ Answered by mrW1 last updated on 14/Dec/17 $$=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\mathrm{sin}\:{x}\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:{x}\right)}\:{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\mathrm{ln}\:\frac{\mathrm{1}−\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{1}+\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}+{C} \\…
Question Number 156841 by mnjuly1970 last updated on 16/Oct/21 Answered by gsk2684 last updated on 16/Oct/21 $${let}\:\boldsymbol{{C}}=\mathrm{cos}\:\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{2}{n}+\mathrm{1}}…\mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{cos}\:\frac{{n}\pi}{\mathrm{2}{n}+\mathrm{1}} \\ $$$${let}\:{S}=\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}{n}+\mathrm{1}}…\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{sin}\frac{{n}\pi}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{2}^{{n}} .{S}.{C}=\left(\mathrm{2sin}\:\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\mathrm{cos}\:\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\left(\mathrm{2sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\left(\mathrm{2sin}\:\frac{\mathrm{3}\pi}{\mathrm{2}{n}+\mathrm{1}}\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)…\left(\mathrm{2sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}\mathrm{cos}\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\left(\mathrm{2sin}\:\frac{{n}\pi}{\mathrm{2}{n}+\mathrm{1}}\mathrm{cos}\:\frac{{n}\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{sin}\:\frac{\mathrm{6}\pi}{\mathrm{2}{n}+\mathrm{1}}…\mathrm{sin}\:\frac{\mathrm{2}\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}.\mathrm{sin}\frac{\mathrm{2}{n}\pi}{\mathrm{2}{n}+\mathrm{1}} \\…
Question Number 91308 by jagoll last updated on 29/Apr/20 $$\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{2}}\:{dx}\:? \\ $$ Commented by jagoll last updated on 29/Apr/20 $$\int\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{2}\right)\left({x}^{\mathrm{2}}…
Question Number 25769 by abdo imad last updated on 14/Dec/17 $${a}−{nser}\:{to}\:{question}\:\mathrm{25765}…{we}\:{put}\:{I}=\int_{\mathrm{0}} ^{\infty} \left({cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{and}\:{J}=\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}…{we}\:{have}\:\mathrm{2}\left({I}+{iJ}\right)=\int_{{R}} {e}^{{ix}^{\mathrm{2}{n}} } \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}}…
Question Number 25765 by abdo imad last updated on 14/Dec/17 $${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left(\:{x}^{\mathrm{2}{n}} \right)^{} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx} \\ $$ Terms…
Question Number 25762 by abdo imad last updated on 14/Dec/17 $${find}\:{the}\:{value}\:{of}\int_{\mathrm{0}} ^{\infty} \:{artan}\left(\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}_{} } \:\:{the}\:{key}\:{of}\:{slution}\:{put}\:{F}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:{artan}\left({xt}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \:\:\:{find}\:\partial{F}/\partial{t}\:{first}\:{then}\:{F}\left({t}\right)\:{and}\:{take}\:{t}=\mathrm{2}\:{you}\:{will}\:{of}\:{find}\:{find}\:{the}\:{value}\:{of}\:{integral}.. \\ $$ Terms of Service…