Question Number 88071 by ar247 last updated on 08/Apr/20 Commented by ar247 last updated on 08/Apr/20 $${help}\:{please} \\ $$ Commented by jagoll last updated on…
Question Number 88069 by Power last updated on 08/Apr/20 Commented by Power last updated on 08/Apr/20 $$\lfloor\mathrm{x}\rfloor−\mathrm{greatest}\:\mathrm{integer}\: \\ $$$$\mathrm{x}=\lfloor\mathrm{x}\rfloor+\left\{\mathrm{x}\right\}\:\:\:\:\:\:\:\mathrm{0}\leqslant\left\{\mathrm{x}\right\}<\mathrm{1} \\ $$ Commented by mathmax by…
Question Number 88064 by jagoll last updated on 08/Apr/20 $$\int\:\frac{\mathrm{dx}}{\mathrm{cos}\:\mathrm{x}\left(\mathrm{2}+\mathrm{sin}\:\mathrm{x}\right)}? \\ $$ Answered by john santu last updated on 08/Apr/20 $$\int\:\frac{\mathrm{cos}\:\mathrm{x}\:\mathrm{dx}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{2}+\mathrm{sin}\:\mathrm{x}\right)}\:=\:\int\frac{\mathrm{d}\left(\mathrm{sin}\:\mathrm{x}\right)}{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{2}+\mathrm{sin}\:\mathrm{x}\right)} \\ $$$$\left[\:\mathrm{let}\:\mathrm{2}+\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{t}\:\right]…
Question Number 88045 by jagoll last updated on 08/Apr/20 $$\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$ Commented by john santu last updated on 08/Apr/20 $$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}\:=\:\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{5}}\left[\frac{\mathrm{2cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{2sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}\:\right] \\ $$$$\Rightarrow\:\int\:\left\{\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{5}}\left[\frac{\mathrm{2cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{2sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}\:\right]\right\}\:\mathrm{dx} \\ $$$$=\:\frac{\mathrm{2x}}{\mathrm{5}}−\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{log}\:\mid\:\mathrm{2sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\:\mid\:+\:\mathrm{c}…
Question Number 88042 by jagoll last updated on 08/Apr/20 $$\mathrm{find}\:\mathrm{max}\:\mathrm{and}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{5}}{−\mathrm{3cos}\:\mathrm{x}−\mathrm{4sin}\:\mathrm{x}} \\ $$ Answered by mr W last updated on 08/Apr/20 $$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{5}}{−\mathrm{3cos}\:\mathrm{x}−\mathrm{4sin}\:\mathrm{x}} \\ $$$$=\:\frac{\mathrm{1}}{−\left(\frac{\mathrm{3}}{\mathrm{5}}\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{4}}{\mathrm{5}}\mathrm{sin}\:\mathrm{x}\right)}…
Question Number 88033 by M±th+et£s last updated on 07/Apr/20 $${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$ Commented by mathmax by abdo last updated on 08/Apr/20 $${approximate}\:{value}\:{we}\:{have}\:{sinx}\:=\sum_{{n}=\mathrm{0}} ^{\infty}…
Question Number 88026 by Chi Mes Try last updated on 07/Apr/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153555 by mnjuly1970 last updated on 08/Sep/21 $$ \\ $$$$\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}{x}\right).{ln}\left({sin}\left({x}\right)\right){dx}\overset{?} {=}\:−\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{solution}\:\left(\mathrm{1}\:\right) \\ $$$$\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left(\:\mathrm{2}{cos}^{\:\mathrm{2}} \left({x}\right)−\mathrm{1}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\::=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}}…
Question Number 88010 by Chi Mes Try last updated on 07/Apr/20 $$\int\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$ Commented by niroj last updated on 07/Apr/20 $$\:\:\:\int\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}}{\:\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 88007 by Mikael_786 last updated on 07/Apr/20 $$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dx}=? \\ $$ Answered by MJS last updated on 07/Apr/20 $${t}=\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}\:\rightarrow\:{dx}=−{x}^{\mathrm{2}} \sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}\sqrt{\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}−\mathrm{1}}{dt} \\ $$$$\Rightarrow…