Question Number 213871 by 073 last updated on 19/Nov/24
Commented by Frix last updated on 20/Nov/24
$$\mathrm{Elliptic}\:\mathrm{Integral}: \\ $$$$\mathrm{4}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\sqrt{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}\:+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{x}}\:{dx}\:=\mathrm{4}{a}\mathrm{E}\:\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$
Commented by BHOOPENDRA last updated on 20/Nov/24
https://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_second_kind
Commented by 073 last updated on 20/Nov/24
$${solution}?? \\ $$
Commented by 073 last updated on 20/Nov/24
$${solution}?? \\ $$$$ \\ $$
Commented by Frix last updated on 20/Nov/24
$$\mathrm{This}\:{is}\:\mathrm{the}\:\mathrm{solution}.\:\mathrm{It}'\mathrm{s}\:\mathrm{defined}\:\mathrm{like}\:\mathrm{this}. \\ $$$$\sqrt{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:{x}\:+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:{x}}=\sqrt{{a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}^{\mathrm{2}} \:{x}}= \\ $$$$={a}\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{a}\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{sin}^{\mathrm{2}} \:{x}}\:{dx}\:={a}\mathrm{E}\:\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\mathrm{Search}\:\mathrm{the}\:\mathrm{www}\:\mathrm{for}\:\mathrm{Elliptic}\:\mathrm{Functions}. \\ $$