Question Number 85718 by M±th+et£s last updated on 24/Mar/20 $$\int\frac{{sin}\left({x}\right)−{cos}\left(\mathrm{3}{x}\right)}{{sin}\left({x}\right)−{cos}\left(\mathrm{2}{x}\right)}{dx} \\ $$ Answered by MJS last updated on 24/Mar/20 $$\int\frac{\mathrm{sin}\:{x}\:−\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{sin}\:{x}\:−\mathrm{cos}\:\mathrm{2}{x}}{dx}= \\ $$$$=\int\frac{\mathrm{4cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{−\mathrm{2cos}^{\mathrm{2}} \:{x}\:+\mathrm{sin}\:{x}\:+\mathrm{1}}{dx}= \\…
Question Number 20182 by DKumar last updated on 23/Aug/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85717 by M±th+et£s last updated on 24/Mar/20 $$\int_{\mathrm{0}} ^{\mathrm{2}} {x}^{\mathrm{4}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{10}} \left(\mathrm{1}−{x}^{{n}} \right){dx} \\ $$$$ \\…
Question Number 151248 by mathdanisur last updated on 19/Aug/21 Answered by EDWIN88 last updated on 19/Aug/21 $$\mathrm{4sin}\:\left(\mathrm{4}{x}−\mathrm{60}°\right)\mathrm{sin}\:\left(\mathrm{6}{x}−\mathrm{60}°\right)\mathrm{sin}\left(\mathrm{480}°−\mathrm{10}{x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left\{\mathrm{2sin}\:\left(\mathrm{4}{x}−\mathrm{60}°\right)\mathrm{sin}\:\left(\mathrm{6}{x}−\mathrm{60}°\right)\right\}\mathrm{2sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\left\{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\left(\mathrm{10}{x}−\mathrm{120}°\right)\right\}\mathrm{2sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\mathrm{2cos}\:\mathrm{2}{x}\:\mathrm{sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)−\mathrm{sin}\:\left(\mathrm{240}°−\mathrm{20}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\mathrm{120}°−\mathrm{8}{x}\right)−\mathrm{sin}\:\left(\mathrm{12}{x}−\mathrm{120}°\right)+\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{20}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0}…
Question Number 20177 by DKumar last updated on 23/Aug/17 $${But}\:{ans}\:{is}\:\frac{\mathrm{1}}{\mathrm{108}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85711 by Rio Michael last updated on 24/Mar/20 $$\:\underset{−\mathrm{4}} {\overset{\mathrm{2}} {\int}}\:\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{\left({x}^{\mathrm{2}} +\:{x}\:+\:\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }\:{dx} \\ $$ Commented by jagoll last updated on 24/Mar/20 $$\underset{−\mathrm{4}}…
Question Number 20174 by DKumar last updated on 23/Aug/17 $$\left(\mathrm{0}.\overset{−} {\mathrm{1}}\right)^{\mathrm{2}} \left\{\mathrm{1}−\mathrm{9}\left(\mathrm{0}.\mathrm{1}\overset{−} {\mathrm{6}}\right)^{\mathrm{2}} \right\} \\ $$ Answered by mrW1 last updated on 23/Aug/17 $$\mathrm{0}.\overline {\mathrm{1}}=\mathrm{0}.\mathrm{111111}…=\frac{\mathrm{1}}{\mathrm{9}}…
Question Number 151247 by mathdanisur last updated on 19/Aug/21 $$\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} \centerdot\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} \centerdot\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{4}} \centerdot\mathrm{4}}\:+\:…\:=\:? \\ $$ Answered by qaz last updated on 19/Aug/21 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n5}^{\mathrm{n}}…
Question Number 85708 by Roland Mbunwe last updated on 24/Mar/20 $${li}\underset{{n}−\infty} {{m}}\:\:/\frac{{U}_{{n}} +\mathrm{1}}{{Un}}/\:\:\:>\mathrm{0} \\ $$$${Test}\:{for}\:{convergence} \\ $$ Answered by Rio Michael last updated on 24/Mar/20…
Question Number 151246 by mathmax by abdo last updated on 19/Aug/21 $$\mathrm{find}\:\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx}\:\mathrm{and}\:\mathrm{J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx} \\ $$ Answered by qaz last updated on 19/Aug/21…