Question Number 65022 by ajfour last updated on 24/Jul/19 Commented by MJS last updated on 24/Jul/19 $$\mathrm{I}\:\mathrm{used}\:\mathrm{a}\:\mathrm{calculator}.\:\mathrm{I}'\mathrm{ve}\:\mathrm{got}\:\mathrm{an}\:\mathrm{old}\:\mathrm{TI}−\mathrm{89} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{any}\:\mathrm{calculator}\:\mathrm{which}\:\mathrm{is}\:\mathrm{able}\:\mathrm{to} \\ $$$$\mathrm{approximately}\:\mathrm{solve}\:\mathrm{equations}\:\mathrm{and}\:\mathrm{integrals} \\ $$$$\mathrm{select}\:\mathrm{a}\:\mathrm{value}\:\mathrm{for}\:{r} \\ $$$$\mathrm{calculate}\:{P},\:{Q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{integrals}…
Question Number 130556 by Ar Brandon last updated on 26/Jan/21 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{xcosx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 130557 by mohammad17 last updated on 26/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 130555 by Dwaipayan Shikari last updated on 26/Jan/21 $$\frac{\mathrm{1}}{\left(\mathrm{2}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}+\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{6}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{6}+\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{10}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{10}+\pi\right)^{\mathrm{2}} }+..=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}{sec}^{\mathrm{2}} \left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$${Prove}\:{or}\:{disprove} \\ $$ Answered…
Question Number 130553 by Study last updated on 26/Jan/21 $${log}\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+….\right)=? \\ $$ Answered by Dwaipayan Shikari last updated on 26/Jan/21 $${log}\left(−{log}\left(\mathrm{1}−{x}\right)\right)={log}\left({log}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\right) \\…
Question Number 65015 by aliesam last updated on 24/Jul/19 $$\int\frac{\sqrt{{x}+\mathrm{1}}\:−\:\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}\:+\:\sqrt{{x}−\mathrm{1}}}\:{dx} \\ $$ Commented by mathmax by abdo last updated on 24/Jul/19 $${let}\:{I}\:=\int\:\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}}{dx}\:\Rightarrow{I}\:=\int\:\frac{\left(\sqrt{{x}+\mathrm{1}}−\sqrt{{x}−\mathrm{1}}\right)^{\mathrm{2}} }{{x}+\mathrm{1}−{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}^{\mathrm{2}}…
Question Number 65013 by Rio Michael last updated on 24/Jul/19 $${why}\:{do}\:{we}\:{divide}\:{each}\:{term}\:{by}\:{n}\:{when}\:{given}\:{the}\:{question} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}\:+\mathrm{2}{n}}{\mathrm{1}+{n}}\:? \\ $$ Answered by Tanmay chaudhury last updated on 24/Jul/19 $$\underset{{n}\rightarrow\infty}…
Question Number 130549 by EDWIN88 last updated on 26/Jan/21 $${the}\:{solution}\:{of}\:{equation}\: \\ $$$$\mid{z}\mid−{z}\:=\:\mathrm{1}+\mathrm{2}{i}\:{is}\:\_\_ \\ $$ Answered by Dwaipayan Shikari last updated on 26/Jan/21 $${z}={x}+{iy} \\ $$$$\sqrt{{x}^{\mathrm{2}}…
Question Number 65011 by AnjanDey last updated on 24/Jul/19 $$\mathrm{1}.\left(\mathrm{i}\right)\mathrm{Evaluate}:\int\frac{\mathrm{1}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}+\sqrt{\mathrm{2}}}{dx} \\ $$$$\left(\mathrm{ii}\right)\mathrm{Evaluate}:\int\mathrm{2}^{\mathrm{2}^{\mathrm{2}^{{x}} } } \mathrm{2}^{\mathrm{2}^{{x}} } \mathrm{2}^{{x}} \:{dx} \\ $$$$\left(\mathrm{iii}\right)\mathrm{Evaluate}:\int\frac{\mathrm{cos}\:^{\mathrm{3}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:{x}}{dx} \\ $$$$\mathrm{2}.\mathrm{cosec}\:\left[\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{cos}\:\left(\mathrm{cot}^{−\mathrm{1}}…
Question Number 130544 by EDWIN88 last updated on 26/Jan/21 $$\:\int\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:+\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\mathrm{tan}\:{x}\:\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)}\:{dx}? \\ $$ Answered by mindispower last updated on 26/Jan/21…