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

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

$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta{d}\theta}{\sqrt{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}}\:=\:? \\ $$

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

$$\mathrm{this}\:\left(\mathrm{like}\:\mathrm{some}\:\mathrm{others}\:\mathrm{you}\:\mathrm{posted}\right)\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{an}\:\mathrm{elliptic}\:\mathrm{integral}.\:\mathrm{see}\:\mathrm{wikipedia}\:\mathrm{for}\:\mathrm{more} \\ $$$$\mathrm{info}.\:\mathrm{elliptic}\:\mathrm{integrals}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{to} \\ $$$$\mathrm{elementar}\:\mathrm{functions}.\:\mathrm{it}'\mathrm{s}\:\mathrm{also}\:\mathrm{not}\:\mathrm{possible} \\ $$$$\mathrm{to}\:\mathrm{exactly}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{ellipse} \\ $$

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

$$\frac{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\sqrt{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta+{b}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}}=\frac{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\sqrt{{b}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{tan}^{\mathrm{2}} \theta\right)}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \frac{\frac{{a}^{\mathrm{2}} }{{b}}{sin}\theta{tan}\theta}{\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{tan}^{\mathrm{2}} \theta}}{d}\theta=\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {f}\left(\theta\right){d}\theta \\ $$$${sin}\left(\mathrm{2}\Pi−\theta\right)=−{sin}\theta \\ $$$${tan}\left(\mathrm{2}\Pi−\theta\right)=−{tan}\theta \\ $$$${sin}\left(\Pi−\theta\right)={sin}\theta \\ $$$${tan}\left(\Pi−\theta\right)=−{tan}\theta \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {f}\left(\theta\right){d}\theta=\mathrm{2}\int_{\mathrm{0}} ^{\Pi} {f}\left(\theta\right){d}\theta\:\:\:\left\{{sincef}\left(\mathrm{2}\Pi−\theta\right)={f}\left(\theta\right)\right\} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\Pi} {f}\left(\theta\right){d}\theta=\mathrm{2}\int_{\mathrm{0}} ^{\Pi} {f}\left(\Pi−\theta\right){d}\theta=−\mathrm{2}\int_{\mathrm{0}} ^{\Pi} {f}\left(\theta\right){d}\theta \\ $$$$\mathrm{4}\int_{\mathrm{0}} ^{\Pi} {f}\left(\theta\right){d}\theta=\mathrm{0} \\ $$$${so}\:{given}\:{intregal}\:{value}\:{is}\:{zero} \\ $$$$ \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\Pi} \frac{\frac{{a}^{\mathrm{2}} }{{b}}{sin}\theta{tan}\theta}{\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{tan}^{\mathrm{2}} \theta}}{d}\theta=\mathrm{2}\int_{\mathrm{0}} ^{\Pi} \frac{\frac{{a}^{\mathrm{2}} }{{b}}{sin}\left(\Pi−\theta\right){tan}\left(\Pi−\theta\right)}{\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{tan}^{\mathrm{2}} \left(\Pi−\theta\right)}}{d}\theta \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\Pi} \frac{\frac{{a}^{\mathrm{2}} }{{b}}×−{sin}\theta{tan}\theta}{\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{tan}^{\mathrm{2}} \theta}}{d}\theta \\ $$$${so}\:\mathrm{4}\int_{\mathrm{0}} ^{\Pi} \frac{\frac{{a}^{\mathrm{2}} }{{b}}{sin}\theta{tan}\theta}{\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{tan}^{\mathrm{2}} \theta}}{d}\theta=\mathrm{0} \\ $$

Commented by ajfour last updated on 13/Jun/18

should not be !

$${should}\:{not}\:{be}\:! \\ $$

Commented by ajfour last updated on 13/Jun/18

((a^2 sin^2 θ)/(√(b^2 cos^2 θ(1+(a^2 /b^2 )tan^2 θ))))  =((a^2 sin^2 θ)/(∣bcos θ∣(√(1+(a^2 /b^2 )tan^2 θ))))  =((a^2 ∣sin θ tan θ∣)/(b(√(1+(a^2 /b^2 )tan^2 θ))))  .....

$$\frac{{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\sqrt{{b}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{tan}^{\mathrm{2}} \theta\right)}} \\ $$$$=\frac{{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{\mid{b}\mathrm{cos}\:\theta\mid\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\mathrm{tan}\:^{\mathrm{2}} \theta}} \\ $$$$=\frac{{a}^{\mathrm{2}} \mid\mathrm{sin}\:\theta\:\mathrm{tan}\:\theta\mid}{{b}\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\mathrm{tan}\:^{\mathrm{2}} \theta}} \\ $$$$..... \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18

in 4th quadrant sinθ and tanθ both negative  in 2nd sinθ=+ve  tanθ=−ve...why mod sign  ∣sinθtanθ∣ come here...

$${in}\:\mathrm{4}{th}\:{quadrant}\:{sin}\theta\:{and}\:{tan}\theta\:{both}\:{negative} \\ $$$${in}\:\mathrm{2}{nd}\:{sin}\theta=+{ve}\:\:{tan}\theta=−{ve}...{why}\:{mod}\:{sign} \\ $$$$\mid{sin}\theta{tan}\theta\mid\:{come}\:{here}... \\ $$

Commented by ajfour last updated on 13/Jun/18

(√x^2 ) =∣x∣  in 2nd quadrant  ∣sin θ tan θ∣=−sin θ tan θ

$$\sqrt{{x}^{\mathrm{2}} }\:=\mid{x}\mid \\ $$$${in}\:\mathrm{2}{nd}\:{quadrant} \\ $$$$\mid\mathrm{sin}\:\theta\:\mathrm{tan}\:\theta\mid=−\mathrm{sin}\:\theta\:\mathrm{tan}\:\theta \\ $$

Answered by ajfour last updated on 14/Jun/18

l=2π(√(x^2 +y^2 ))     =2π(√(a^2 +((a^2 b^2 )/a^2 ))) = 2π(√(a^2 +b^2 )) .  Why isn′t this correct if above  integral is same as perimeter of  ellipse.

$${l}=\mathrm{2}\pi\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\:\:\:=\mathrm{2}\pi\sqrt{{a}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:=\:\mathrm{2}\pi\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:. \\ $$$${Why}\:{isn}'{t}\:{this}\:{correct}\:{if}\:{above} \\ $$$${integral}\:{is}\:{same}\:{as}\:{perimeter}\:{of} \\ $$$${ellipse}.\: \\ $$

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