I decided to solve it for $[0, \pi/2]$ as a first thing. $(2a\sin^2{t})' = 2a\sin{2t}$ and $(2a\cos{t})' = -2a\sin{t}$. So $$l = \int_0^{\pi/2}{{2a\sqrt{4\sin^2{t}\cos^2{t} + \sin^2{t}}\,dt} = 2a\int_0^{\pi/2}{\sin{t}\sqrt{1 + 4\cos^2{t}}}\,dt}$$ Now I go with substitution. $u = \cos{t}$. $$2a\int_0^1{\dfrac{\sqrt{1-u^2}\sqrt{1+4u^2}}{\sqrt{1-u^2}}du} = 2a\int_0^1 {\sqrt{1+4u^2}du}$$ From here on I can find the antiderivative of the integrand and apply the fundamental formula of calculus and solve it. Actually, if I check its antiderivative using an online integral calculator and then apply the formula and finally multiply by 4 (because I have to solve it for $0 \leq t \leq 2\pi$) then I'll get the right answer. But is there an easier solution? This problem is marked as easy and usually such problems in this textbook are solved in 3-4 steps.
2026-03-28 14:00:29.1774706429
Arc length of curve $x = 2a\sin^2{t}$, $y = 2a\cos{t}$ $0 \leq t \leq 2\pi$. Easier solution?
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