Question Number 208540 by otchereabdullai@gmail.com last updated on 17/Jun/24 Answered by Sutrisno last updated on 18/Jun/24 $${misal}\::\:\mathrm{3}{x}=\mathrm{2}{tan}\theta \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta\rightarrow{dx}=\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\int\frac{\frac{\mathrm{2}}{\mathrm{3}}{tan}\theta−\mathrm{3}}{\:\sqrt{\left(\mathrm{2}{tan}\theta\right)^{\mathrm{2}} +\mathrm{4}}}.\frac{\mathrm{2}}{\mathrm{3}}{sec}^{\mathrm{2}} \theta{d}\theta \\…
Question Number 208502 by Kalebwizeman last updated on 17/Jun/24 Answered by A5T last updated on 17/Jun/24 $$=\sqrt{\mathrm{16}}+\sqrt{\mathrm{15}}−\sqrt{\mathrm{15}}−\sqrt{\mathrm{14}}+…−\sqrt{\mathrm{9}}−\sqrt{\mathrm{8}}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$ Commented by Kalebwizeman last updated on…
Question Number 208513 by CrispyXYZ last updated on 17/Jun/24 $${a},\:{b}>\mathrm{0}.\:\mathrm{2}{a}+{b}=\mathrm{1},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}{a}}+\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} }\:>\:\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208515 by mokys last updated on 17/Jun/24 $$\int_{\mathrm{0}} ^{\:\infty} \:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208440 by liuxinnan last updated on 16/Jun/24 $${a}_{\mathrm{1}} >{a}_{\mathrm{2}} >{a}_{\mathrm{3}} >…>{a}_{{n}} >\mathrm{0} \\ $$$${b}_{\mathrm{1}} >{b}_{\mathrm{2}} >{b}_{\mathrm{3}} >…>{b}_{{n}} >\mathrm{0} \\ $$$${prove} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}}…
Question Number 208398 by mokys last updated on 14/Jun/24 $${write}\:{z}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+{i}}\:{in}\:{e}^{{i}\theta} \\ $$ Answered by A5T last updated on 14/Jun/24 $${z}=\frac{\sqrt{\mathrm{3}}−{i}}{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left({i}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}−{i}}{\mathrm{4}},\:\mid{z}\mid=\sqrt{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\…
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Question Number 208354 by kgmxdd last updated on 13/Jun/24 Commented by Frix last updated on 13/Jun/24 $$=\frac{\mathrm{4}}{\pi}!!! \\ $$$$\left(\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{online}.\right) \\ $$ Terms of Service Privacy…
Question Number 208338 by behnamraygan last updated on 12/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208318 by SANOGO last updated on 11/Jun/24 $${calcul}\:\:\:{lim}\:{n}\rightarrow+\infty \\ $$$$\int_{\mathrm{0}} ^{+\infty} \:\frac{{cos}\left({nx}\right)}{\left({nx}+\mathrm{1}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:}{dx} \\ $$ Answered by Berbere last updated on 11/Jun/24 $$\mid\int_{\mathrm{0}}…