Question Number 145021 by mim24 last updated on 01/Jul/21 Answered by phally last updated on 01/Jul/21 $$\:\mathrm{Derivative}\:\mathrm{formula} \\ $$$$\:\left(\mathrm{Cos}^{−\mathrm{1}} \left(\mathrm{u}\right)\right)^{'} =\frac{−\mathrm{u}'}{\:\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }} \\ $$ Answered…
Question Number 79480 by TawaTawa last updated on 25/Jan/20 Commented by mathmax by abdo last updated on 25/Jan/20 $${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left({tanx}\right)}{{cos}\left(\mathrm{2}{x}\right)}{dx}\:\:{changement}\:{tanx}\:={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\frac{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 144995 by liberty last updated on 01/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{distance}\: \\ $$$$\mathrm{between}\:\mathrm{two}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\:\mathrm{curve}\:\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{a}^{\mathrm{4}} }\:+\:\frac{\mathrm{y}^{\mathrm{4}} }{\mathrm{b}^{\mathrm{4}} }\:=\:\mathrm{1}\:. \\ $$ Answered by mr W last…
Question Number 13871 by Nayon last updated on 24/May/17 $${why}\:{the}\:{function},{sin}\left({x}\right)\:{is}\:{a}\:{power} \\ $$$${series}?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 79398 by ubaydulla last updated on 24/Jan/20 $$\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$ Commented by john santu last updated on 24/Jan/20 $$\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:\int\:\mathrm{1}\:{dx}−\int\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}}…
Question Number 144899 by mnjuly1970 last updated on 30/Jun/21 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\Omega\::=\:\int\:\frac{\sqrt{\mathrm{1}−{sin}\left({x}\right)}}{{cos}\:\left({x}\right)}\:{e}\:^{−\frac{\mathrm{1}}{\mathrm{2}}\:{x}} =\:? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 13806 by ajfour last updated on 23/May/17 $${Prove}\:{that}\:{for}\:−\frac{\pi}{\mathrm{2}}<{x}<\frac{\pi}{\mathrm{2}}\:, \\ $$$$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\mathrm{cos}\:\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }\mathrm{cos}\:\mathrm{5}{x}−….{to}\:{infinity} \\ $$$$\:\:=\frac{\pi}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−{x}^{\mathrm{2}} \right)\:. \\ $$ Commented by prakash jain…
Question Number 13751 by Nayon last updated on 23/May/17 $$\frac{{ds}}{{dt}}={v},\frac{{dv}}{{dt}}={a},\frac{{da}}{{dt}}={b},\frac{{db}}{{dt}}={e},\frac{{de}}{{dt}}={f} \\ $$$$\frac{{df}}{{dt}}={g},\frac{{dg}}{{dt}}={h},\frac{{dh}}{{dt}}={i},\frac{{di}}{{dt}}={j},\frac{{dj}}{{dt}}={k},….. \\ $$$${now}\:{if}\:{we}\:{continue}\:{this}\:{process}\:{to} \\ $$$${infinity}..{and}\:{if}\:{v}_{\mathrm{0}} ,{v},{a},{b},{e},{f},{g},{h},{i}, \\ $$$${j},…………….=\mathrm{1}\:.{then}\:{calculate} \\ $$$${the}\:{formula}\:{of}\:{v}\:{and}\:{s}\:… \\ $$$$ \\ $$…
Question Number 144684 by mnjuly1970 last updated on 27/Jun/21 $$ \\ $$ Answered by mindispower last updated on 27/Jun/21 $${M}:={xyz}−\left({xy}+{yz}+{zx}\right)+{x}+{y}+{z}−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\mathrm{1}{xyz}={xy}+{yz}+{zx} \\ $$$${M}={x}+{y}+{z}−\mathrm{1} \\…
Question Number 79126 by ~blr237~ last updated on 22/Jan/20 $${Study}\:\:\:{f}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{x}^{{n}} {sin}\left({nx}\right)}{{n}} \\ $$$${Find}\:{out}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\frac{{sin}\left({n}\right)}{{n}}\:\:\:{and}\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\right)}{{n}}\: \\ $$ Commented by mathmax…