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Question Number 142708 by ajfour last updated on 05/Jun/21

I=∫_0 ^( α) (√(c^2 −sin^2 θ))dθ tan α=(a^2 /b^2 ) , a^2 >b^2 , c^2 >1 Perimeter of ellipse =4∫_0 ^( π/2) (√(a^2 −(a^2 −b^2 )sin^2 θ)) dθ (is that right sir?)

$$\:{I}=\int_{\mathrm{0}} ^{\:\:\alpha} \sqrt{{c}^{\mathrm{2}} −\mathrm{sin}\:^{\mathrm{2}} \theta}{d}\theta \\ $$ $$\:\mathrm{tan}\:\alpha=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\:,\:{a}^{\mathrm{2}} >{b}^{\mathrm{2}} \:\:,\:{c}^{\mathrm{2}} >\mathrm{1} \\ $$ $${Perimeter}\:{of}\:{ellipse} \\ $$ $$=\mathrm{4}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \sqrt{{a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:^{\mathrm{2}} \theta}\:{d}\theta \\ $$ $$\left({is}\:{that}\:{right}\:{sir}?\right) \\ $$

Answered by MJS_new last updated on 04/Jun/21

∫(√(c^2 −sin^2 θ)) dθ=c∫(√(1−c^(−2) sin^2 θ)) dθ= =cE (θ∣c^(−2) ) +C

$$\int\sqrt{{c}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\theta}\:{d}\theta={c}\int\sqrt{\mathrm{1}−{c}^{−\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}\:{d}\theta= \\ $$ $$={c}\mathrm{E}\:\left(\theta\mid{c}^{−\mathrm{2}} \right)\:+{C} \\ $$

Commented byajfour last updated on 05/Jun/21

whats the name?

$${whats}\:{the}\:{name}? \\ $$

Commented byMJS_new last updated on 05/Jun/21

Incomplete Elliptic Integral of the 2^(nd) Kind

$$\mathrm{Incomplete}\:\mathrm{Elliptic}\:\mathrm{Integral}\:\mathrm{of}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{Kind} \\ $$

Commented bymohammad17 last updated on 05/Jun/21

sir can you give me the formola

$${sir}\:{can}\:{you}\:{give}\:{me}\:{the}\:{formola} \\ $$

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