∫_0 ^( 2π) ((a^2 sin^2 θdθ)/(√(a^2 sin^2 θ+b^2 cos^2 θ))) = ?


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Question Number 37451 by ajfour last updated on 13/Jun/18

$\int_{0}^{2 π} \frac{a^{2} \sin^{2} θ d θ}{\sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}} = ?$
Commented by MJS last updated on 13/Jun/18

$this (like some others you posted) leads to$ $an elliptic integral . see wikipedia for more$ $info . elliptic integrals cannot be solved to$ $elementar functions . {it}^{'} s also not possible$ $to exactly calculate the perimeter of an$ $ellipse$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
	$((a^2 sin^2 θ)/(√(a^2 sin^2 θ+b^2 cos^2 θ)))=((a^2 sin^2 θ)/(√(b^2 cos^2 θ(1+(a^2 /b^2 )tan^2 θ)))) ∫_0 ^(2Π) (((a^2 /b)sinθtanθ)/(√(1+(a^2 /b^2 )tan^2 θ)))dθ=∫_0 ^(2Π) f(θ)dθ sin(2Π−θ)=−sinθ tan(2Π−θ)=−tanθ sin(Π−θ)=sinθ tan(Π−θ)=−tanθ ∫_0 ^(2Π) f(θ)dθ=2∫_0 ^Π f(θ)dθ {sincef(2Π−θ)=f(θ)} 2∫_0 ^Π f(θ)dθ=2∫_0 ^Π f(Π−θ)dθ=−2∫_0 ^Π f(θ)dθ 4∫_0 ^Π f(θ)dθ=0 so given intregal value is zero 2∫_0 ^Π (((a^2 /b)sinθtanθ)/(√(1+(a^2 /b^2 )tan^2 θ)))dθ=2∫_0 ^Π (((a^2 /b)sin(Π−θ)tan(Π−θ))/(√(1+(a^2 /b^2 )tan^2 (Π−θ))))dθ =−2∫_0 ^Π (((a^2 /b)×−sinθtanθ)/(√(1+(a^2 /b^2 )tan^2 θ)))dθ so 4∫_0 ^Π (((a^2 /b)sinθtanθ)/(√(1+(a^2 /b^2 )tan^2 θ)))dθ=0$
	$\frac{a^{2} {s i n}^{2} θ}{\sqrt{a^{2} {s i n}^{2} θ + b^{2} {c o s}^{2} θ}} = \frac{a^{2} {s i n}^{2} θ}{\sqrt{b^{2} {c o s}^{2} θ (1 + \frac{a^{2}}{b^{2}} {t a n}^{2} θ)}}$ $\int_{0}^{2 Π} \frac{\frac{a^{2}}{b} s i n θ t a n θ}{\sqrt{1 + \frac{a^{2}}{b^{2}} {t a n}^{2} θ}} d θ = \int_{0}^{2 Π} f (θ) d θ$ $s i n (2 Π - θ) = - s i n θ$ $t a n (2 Π - θ) = - t a n θ$ $s i n (Π - θ) = s i n θ$ $t a n (Π - θ) = - t a n θ$ $\int_{0}^{2 Π} f (θ) d θ = 2 \int_{0}^{Π} f (θ) d θ {s i n c e f (2 Π - θ) = f (θ)}$ $2 \int_{0}^{Π} f (θ) d θ = 2 \int_{0}^{Π} f (Π - θ) d θ = - 2 \int_{0}^{Π} f (θ) d θ$ $4 \int_{0}^{Π} f (θ) d θ = 0$ $s o g i v e n i n t r e g a l v a l u e i s z e r o$ $2 \int_{0}^{Π} \frac{\frac{a^{2}}{b} s i n θ t a n θ}{\sqrt{1 + \frac{a^{2}}{b^{2}} {t a n}^{2} θ}} d θ = 2 \int_{0}^{Π} \frac{\frac{a^{2}}{b} s i n (Π - θ) t a n (Π - θ)}{\sqrt{1 + \frac{a^{2}}{b^{2}} {t a n}^{2} (Π - θ)}} d θ$ $= - 2 \int_{0}^{Π} \frac{\frac{a^{2}}{b} \times - s i n θ t a n θ}{\sqrt{1 + \frac{a^{2}}{b^{2}} {t a n}^{2} θ}} d θ$ $s o 4 \int_{0}^{Π} \frac{\frac{a^{2}}{b} s i n θ t a n θ}{\sqrt{1 + \frac{a^{2}}{b^{2}} {t a n}^{2} θ}} d θ = 0$
	Commented by ajfour last updated on 13/Jun/18

	$s h o u l d n o t b e!$
	Commented by ajfour last updated on 13/Jun/18

	$\frac{a^{2} {s i n}^{2} θ}{\sqrt{b^{2} {c o s}^{2} θ (1 + \frac{a^{2}}{b^{2}} {t a n}^{2} θ)}}$ $= \frac{a^{2} \sin^{2} θ}{∣ b \cos θ ∣ \sqrt{1 + \frac{a^{2}}{b^{2}} \tan^{2} θ}}$ $= \frac{a^{2} ∣ \sin θ \tan θ ∣}{b \sqrt{1 + \frac{a^{2}}{b^{2}} \tan^{2} θ}}$ $. . . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18

	$i n 4 t h q u a d r a n t s i n θ a n d t a n θ b o t h n e g a t i v e$ $i n 2 n d s i n θ = + v e t a n θ = - v e . . . w h y m o d s i g n$ $∣ s i n θ t a n θ ∣ c o m e h e r e . . .$
	Commented by ajfour last updated on 13/Jun/18

	$\sqrt{x^{2}} =∣ x ∣$ $i n 2 n d q u a d r a n t$ $∣ \sin θ \tan θ ∣= - \sin θ \tan θ$
	Answered by ajfour last updated on 14/Jun/18

	$l = 2 π \sqrt{x^{2} + y^{2}}$ $= 2 π \sqrt{a^{2} + \frac{a^{2} b^{2}}{a^{2}}} = 2 π \sqrt{a^{2} + b^{2}} .$ $W h y {i s n}^{'} t t h i s c o r r e c t i f a b o v e$ $i n t e g r a l i s s a m e a s p e r i m e t e r o f$ $e l l i p s e .$
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