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?)


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

$I = \int_{0}^{α} \sqrt{c^{2} - \sin^{2} θ} d θ$ $\tan α = \frac{a^{2}}{b^{2}}, a^{2} > b^{2}, c^{2} > 1$ $P e r i m e t e r o f e l l i p s e$ $= 4 \int_{0}^{π / 2} \sqrt{a^{2} - (a^{2} - b^{2}) \sin^{2} θ} d θ$ $(i s t h a t r i g h t s i r ?)$
	Answered by MJS_new last updated on 04/Jun/21

	$\int \sqrt{c^{2} - \sin^{2} θ} d θ = c \int \sqrt{1 - c^{- 2} \sin^{2} θ} d θ =$ $= c E (θ ∣ c^{- 2}) + C$
	Commented byajfour last updated on 05/Jun/21

	$w h a t s t h e n a m e ?$
	Commented byMJS_new last updated on 05/Jun/21

	$Incomplete Elliptic Integral of the 2^{nd} Kind$
	Commented bymohammad17 last updated on 05/Jun/21

	$s i r c a n y o u g i v e m e t h e f o r m o l a$
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