I-0-c-2-sin-2-d-tan-a-2-b-2-a-2-gt-b-2-c-2-gt-1-Perimeter-of-ellipse-4-0-pi-2-a-2-a-2-b-2-sin-2-d-is-that-right-sir-

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 by ajfour last updated on 05/Jun/21