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Question-161617

Question Number 161617 by gbanda95 last updated on 20/Dec/21 Answered by mathmax by abdo last updated on 20/Dec/21 $$\mathrm{I}_{\mathrm{n}+\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{n}+\mathrm{2}} \mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}}…

Question-96052

Question Number 96052 by 174 last updated on 29/May/20 Answered by john santu last updated on 29/May/20 $$\left(\mathrm{a}−\mathrm{3}\right)\left(\mathrm{a}+\mathrm{10}\right)\:=\:\mathrm{a}^{\mathrm{2}} +\mathrm{7a}−\mathrm{30} \\ $$$$=\mathrm{13}−\mathrm{30}\:=\:−\mathrm{17}\: \\ $$$$\left(\mathrm{a}+\mathrm{2}\right)\left(\mathrm{a}+\mathrm{5}\right)=\mathrm{a}^{\mathrm{2}} +\mathrm{7a}+\mathrm{10}=\mathrm{23} \\…

sin-p-x-1-

Question Number 96026 by ubaydulla last updated on 29/May/20 $$\mathrm{sin}\:\frac{\mathrm{p}}{\mathrm{x}}=\mathrm{1} \\ $$ Commented by bobhans last updated on 29/May/20 $$\mathrm{sin}\:\left(\frac{\mathrm{p}}{\mathrm{x}}\right)\:=\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{p}}{\mathrm{x}}\:=\:\frac{\pi}{\mathrm{2}}\:+\:\mathrm{2k}\pi\:\Rightarrow\:\frac{\mathrm{x}}{\mathrm{p}}\:=\:\frac{\mathrm{2}}{\pi+\mathrm{4k}\pi}\: \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{2p}}{\pi+\mathrm{4k}\pi} \\…

1-1-a-a-2-1-a-2-

Question Number 96008 by 6532 last updated on 29/May/20 $$\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{a}}}{\mathrm{a}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }} \\ $$ Commented by john santu last updated on 29/May/20 $$\frac{{a}^{\mathrm{2}} −{a}}{{a}^{\mathrm{4}} −\mathrm{1}}\:=\:\frac{{a}\left({a}−\mathrm{1}\right)}{\left({a}−\mathrm{1}\right)\left({a}+\mathrm{1}\right)\left({a}^{\mathrm{2}}…

Calculate-lim-x-0-1-cos-1-cos-x-x-4-lim-x-0-1-cos-xcos-2xcos-3x-x-2-

Question Number 161409 by LEKOUMA last updated on 17/Dec/21 $${Calculate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{4}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$ Commented by cortano last updated…