Question Number 103515 by bemath last updated on 15/Jul/20 $${given}\:{f}\left({x}\right)\:=\:{f}\left({x}+\frac{\pi}{\mathrm{6}}\right)\:\forall{x}\in\mathbb{R} \\ $$$${if}\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{6}} {\int}}{f}\left({x}\right){dx}\:=\:{T}\:{then}\:\underset{\pi} {\overset{\mathrm{7}\pi/\mathrm{3}} {\int}}{f}\left({x}+\pi\right) \\ $$$${dx}\:? \\ $$ Answered by bramlex last updated…
Question Number 103512 by bemath last updated on 15/Jul/20 $$\int\:\frac{{x}\:{dx}}{\left(\mathrm{cot}\:{x}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} }\:= \\ $$$$\left({a}\right)\:\frac{{x}}{\mathrm{16}}−\frac{{x}\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{32}}−\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$$$\left({b}\right)\:\frac{{x}}{\mathrm{16}}+\frac{{x}\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{32}}−\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$$$\left({c}\right)\:\frac{{x}}{\mathrm{16}}+\frac{{x}\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{64}}+\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$$$\left({d}\right)\frac{{x}}{\mathrm{16}}+\frac{{x}\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{32}}+\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$ Commented by bobhans last…
dx-x-x-1-4-1-a-9-x-1-4-1-18-x-1-4-1-9-c-b-9-x-1-4-1-18-x-1-4-1-9-c-c-9-x-1-4-1-18-x-1-4-1-9-c-d-9-x-
Question Number 103511 by bemath last updated on 15/Jul/20 $$\int\:\frac{{dx}}{\:\sqrt{{x}}\:\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)}\:=\_\_ \\ $$$$\left({a}\right)\:−\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}{\mathrm{18}\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)^{\mathrm{9}} }\:+\:{c}\: \\ $$$$\left({b}\right)\:\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}}{\mathrm{18}\left(\sqrt[{\mathrm{4}}]{{x}}+\mathrm{1}\right)^{\mathrm{9}} }\:+{c} \\ $$$$\left({c}\right)\:−\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}\:−\mathrm{1}}{\mathrm{18}\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)^{\mathrm{9}} }\:+{c} \\ $$$$\left({d}\right)\:\frac{\mathrm{9}\:\sqrt[{\mathrm{4}}]{{x}}+\mathrm{1}}{\mathrm{8}\left(\sqrt[{\mathrm{4}}]{{x}}\:+\mathrm{1}\right)^{\mathrm{9}} }\:+\:{c} \\ $$ Commented…
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Question Number 37961 by prof Abdo imad last updated on 19/Jun/18 $${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}\:} \:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:. \\ $$ Answered by MJS last updated on 20/Jun/18…
Question Number 37938 by Fawomath last updated on 19/Jun/18 Commented by abdo.msup.com last updated on 19/Jun/18 $$\left.\mathrm{1}\right)\:{let}\:{I}_{{p}} =\:\int\:{x}^{\mathrm{2}} \left({n}−{x}\right)^{{p}} \:{dx}\:{by}\:{parts} \\ $$$${I}_{{p}} =\:−\frac{\mathrm{1}}{{p}+\mathrm{1}}{x}^{\mathrm{2}} \left({n}−{x}\right)^{{p}+\mathrm{1}} \:\:+\frac{\mathrm{1}}{{p}+\mathrm{1}}\int\:\:\mathrm{2}{x}\:\:\left({n}−{x}\right)^{{p}+\mathrm{1}}…
Question Number 169000 by Skabetix last updated on 22/Apr/22 Answered by haladu last updated on 23/Apr/22 $$=\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$ Commented by Skabetix last updated on…
Question Number 37922 by gunawan last updated on 19/Jun/18 $${f}\::\:\mathbb{N}\:\rightarrow\:\mathbb{R} \\ $$$${g}\::\:\mathbb{N}\:\rightarrow\:\mathbb{R} \\ $$$${f}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {x}^{{n}} \mathrm{sin}\:{x}\:{dx} \\ $$$${g}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {x}^{{n}} \mathrm{cos}\:{x}\:{dx} \\ $$$$\frac{{f}\left({n}+\mathrm{1}\right)−{f}\left({n}\right)}{{g}\left({n}+\mathrm{1}\right)−{g}\left({n}\right)}=? \\…
Question Number 103454 by bemath last updated on 15/Jul/20 $$\int\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{6}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} }\:{dx}\: \\ $$ Answered by bobhans last updated on 15/Jul/20 $${I}=\int{x}\left({x}+\mathrm{2}{x}^{\mathrm{3}}…
Question Number 37912 by gunawan last updated on 19/Jun/18 $$\mathrm{Evaluate}\::\:\mathrm{the}\:\mathrm{Integral} \\ $$$$\int_{-\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{3}\:\mathrm{cos}\:\theta} {r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta.\:{dr}\:{d}\theta\: \\ $$ Commented by math khazana by…