Question Number 91534 by M±th+et+s last updated on 01/May/20 $$\int_{\mathrm{1}} ^{\infty} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$ Commented by mathmax by abdo last updated on 01/May/20…
Question Number 25965 by Chuks” last updated on 17/Dec/17 $${solve}\:{the}\:{integral} \\ $$$$\int\mathrm{sin}^{\mathrm{4}} \theta{d}\theta \\ $$ Answered by $@ty@m last updated on 17/Dec/17 $$\mathrm{sin}\:\mathrm{3}{x}=\mathrm{3sin}\:{x}−\mathrm{4sin}\:^{\mathrm{3}} {x} \\…
Question Number 25960 by abdo imad last updated on 16/Dec/17 $${answer}\:{to}\:\mathrm{25955}.{we}\:{introduce}\:{the}\:{parametric}\:{function} \\ $$$${F}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:{ln}\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}_{} } \right){t}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{after}\:{verifying}\:{that}\: \\ $$$${F}\:{is}\:{derivable}\:{on}\left[\mathrm{0}.\propto\left[\:\:{we}\:{find}\:\:\:\partial{F}/\partial{t}=\:\:\int_{\mathrm{0}} ^{\infty} \left(\:\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){t}\right)^{−\mathrm{1}} {dx}\right.\right.\right.…
Question Number 25955 by abdo imad last updated on 16/Dec/17 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\int_{\mathrm{0}} ^{\infty} \:\:\boldsymbol{\mathrm{ln}}\left(\mathrm{2}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)^{−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157016 by amin96 last updated on 18/Oct/21 $${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}^{{x}} }+\frac{\mathrm{1}}{\mathrm{4}^{{x}} +\mathrm{1}}\:\: \\ $$$${find}\:\:\:\int_{\mathrm{1}} ^{\mathrm{5}} {f}\left({x}\right){dx}+\int_{−\mathrm{5}} ^{−\mathrm{1}} {f}\left({x}\right){dx} \\ $$ Answered by amin96 last…
Question Number 91479 by Zainal Arifin last updated on 01/May/20 $$\:\underset{−\pi/\mathrm{2}\:} {\overset{\pi/\mathrm{2}} {\int}}\:\sqrt{\mathrm{cos}\:{x}−\mathrm{cos}^{\mathrm{3}} {x}}\:\mathrm{dx}=… \\ $$ Commented by jagoll last updated on 01/May/20 $$\sqrt{\mathrm{cos}\:{x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)}\:=\:\sqrt{\mathrm{cos}\:{x}}\:\sqrt{\mathrm{sin}\:^{\mathrm{2}}…
Question Number 25932 by abdo imad last updated on 16/Dec/17 $${answer}\:{to}\:\mathrm{25824}\:\:{we}\:{have}\:{a}^{−{x}^{\mathrm{2}} } \:=\:{e}^{−{x}^{\mathrm{2}_{} } {ln}\left({a}\right)} \:\:{so}\:{for}\:{a}>\mathrm{1} \\ $$$${ln}\left({a}\right)=\left(\:\left({ln}\left({a}\right)\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} >>>>\int_{{R}} {a}^{−^{} {x}^{\mathrm{2}} } \:=\:\int_{{R}} {e}^{−\left({x}\:\left({ln}\left({a}\right)^{\mathrm{1}/\mathrm{2}}…
Question Number 25887 by shivram198922@gmail.com last updated on 16/Dec/17 Answered by ajfour last updated on 16/Dec/17 $$\int_{\mathrm{0}} ^{\:\:\pi} \left[\mathrm{cos}\:\left({p}−{n}\right){x}−\mathrm{cos}\:\left({p}+{n}\right){x}\right]{dx} \\ $$$$=\frac{\mathrm{sin}\:\left({p}−{n}\right)\pi}{{p}−{n}}−\frac{\mathrm{sin}\:\left({p}+{n}\right)\pi}{{p}+{n}} \\ $$$$=\frac{{p}\left[\mathrm{sin}\:\left({p}−{n}\right)\pi−\mathrm{sin}\:\left({p}+{n}\right)\pi\right]}{{p}^{\mathrm{2}} −{n}^{\mathrm{2}} }…
Question Number 25868 by ajfour last updated on 16/Dec/17 Commented by ajfour last updated on 16/Dec/17 $${Find}\:{the}\:{area}\:{enclosed}\:{by}\:{a} \\ $$$${variable}\:{line}\:\frac{{x}}{{t}}+\frac{{y}}{{l}−{t}}=\mathrm{1}\:{and} \\ $$$${the}\:{coordinate}\:{axes}\:{in}\:{the} \\ $$$${first}\:{quadrant};\: \\ $$$$\left({the}\:{line}\:{shifts}\:{as}\:{t}\:{changes}\right.…
Question Number 25865 by anonymous_ last updated on 15/Dec/17 $$\int\frac{\mathrm{2}+\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$ Answered by ajfour last updated on 16/Dec/17 $$\frac{\mathrm{2}+\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}+\mathrm{2}{x}={A}\left({x}^{\mathrm{2}}…