I have a curve defined by parametric equations $$y=a\sin^{5}t \\ x=a\cos^{5}t, $$ where $t\in(0;2\pi)$. I solve it by well-known formula: $L=\int\limits_{0}^{2\pi}\sqrt{(y')^2+(x')^2}dt$. $$x'=-5a\cos^{4}t\cdot\sin t \\ y'=5a\sin^{4}t \cdot\cos t$$
$L=\int\limits_{0}^{2\pi}\sqrt{25a^2\sin^8t \cos^2t+25a^2\cos^8t \sin^2t}dt=\int\limits_{0}^{2\pi}\sqrt{25a^2\sin^2t \cos^2t(\cos^6t + \sin^6t)}dt=5a\int\limits_{0}^{2\pi}\sin t \cos t \sqrt{\sin^6t+\cos^6t}.$
If I will use that $\sin^6t+\cos^6t=1-3\sin^2t+3\sin^4t$, than by using substitute $u=\sin^2t$ I get $L=\frac{5a}{2}\int\limits_{}^{}\sqrt{1-3u-u^2}du$, which can't be solved by standard methods.
If we consider $$I=\int\sqrt{1-3u-u^2}du$$ complete the square $$1-3u-u^2=\frac{13}4-\left(u+\frac 32\right)^2$$ Now, change variable $$u+\frac 32=\frac{ \sqrt{13}}{2} \sin (v)\implies u=\frac{1}{2} \left(\sqrt{13} \sin (v)-3\right)\implies du=\frac{\sqrt{13}}{2} \cos (v)\,dv$$ All of this makes $$I=\frac{13 }{4}\int\cos ^2(v)\,dv$$
I am sure that you can take it from here.