Question Number 17206 by Arnab Maiti last updated on 02/Jul/17 $$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{vallu}\:\mathrm{of}\:\int_{−\mathrm{a}} ^{\:\mathrm{a}} \mathrm{x}^{\mathrm{2}} \mathrm{y}\:\mathrm{dx}\:\:? \\ $$$$\mathrm{Where}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{y}\geqslant\mathrm{0} \\ $$ Answered by mrW1 last…
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Question Number 17205 by Arnab Maiti last updated on 02/Jul/17 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\sqrt{\frac{\mathrm{n}+\mathrm{r}}{\mathrm{n}−\mathrm{r}}} \\ $$ Answered by ajfour last updated on 02/Jul/17 $$\frac{\mathrm{r}}{\mathrm{n}}\rightarrow\mathrm{x}\:,\:\Rightarrow\:\mathrm{dx}\rightarrow\frac{\mathrm{1}}{\mathrm{n}} \\…
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Question Number 17204 by Arnab Maiti last updated on 02/Jul/17 $$\int_{\mathrm{0}} ^{\:\frac{\Pi}{\mathrm{2}}} \mathrm{sin}\theta\:\mathrm{cos}\theta\left(\mathrm{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta+\mathrm{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{d}\theta \\ $$ Answered by prakash jain last…
Question Number 17203 by Arnab Maiti last updated on 02/Jul/17 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$ Commented by prakash jain last updated on 02/Jul/17…
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Question Number 148268 by mathdanisur last updated on 26/Jul/21 Answered by Olaf_Thorendsen last updated on 26/Jul/21 $${z}^{\mathrm{4}} −\mathrm{3}{z}^{\mathrm{2}} +\mathrm{1}\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{4}−{z}^{\mathrm{2}} }−\mathrm{1}}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{w}\:=\:{z}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\::\:{w}^{\mathrm{2}} −\mathrm{3}{w}+\mathrm{1}\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{4}−{w}}−\mathrm{1}}…
Question Number 148270 by abdurehime last updated on 26/Jul/21 $$\mathrm{proof}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{ellipse}\:\mathrm{at} \\ $$$$\mathrm{a}\:\mathrm{center}\left(\mathrm{0}.\mathrm{0}\right)\mathrm{is}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1} \\ $$ Commented by abdurehime last updated on 26/Jul/21…
Question Number 82729 by TawaTawa last updated on 23/Feb/20 Commented by Kunal12588 last updated on 23/Feb/20 $${In}\:{the}\:\bigtriangleup{ABC},\:{let}\:{G}\:{be}\:{the}\:{centroid},\:{and}\:{let}\:{I} \\ $$$${be}\:{the}\:{center}\:{of}\:{the}\:{inscribed}\:{circle}.\:{Let}\:\alpha \\ $$$${and}\:\beta\:{be}\:{the}\:{angles}\:{at}\:{the}\:{vertices}\:{A}\:{and}\:{B} \\ $$$${respectively}.\:{Suppose}\:{that}\:{the}\:{segment}\:{IG}\:\parallel\:{AB} \\ $$$${and}\:{that}\:\beta\:=\:\mathrm{2}\:{tan}^{−\mathrm{1}}…