I am trying to express the eliptic integral in series expression that depends on $a,b,\alpha$ and without integral
$$L(\alpha)=\int_0^\alpha\sqrt{a^2\sin^2 t+b^2 \cos^2 t}\,dt $$
$$\frac {\partial L(\alpha)}{\partial a}=\int_0^\alpha \frac{a\sin^2 t}{\sqrt{a^2\sin^2 t+b^2 \cos^2 t}}\,dt $$
$$\frac {\partial L(\alpha)}{\partial b}=\int_0^\alpha \frac{b\cos^2 t}{\sqrt{a^2\sin^2 t+b^2 \cos^2 t}}\,dt $$
$$a\frac {\partial L(\alpha)}{\partial a}+b\frac {\partial L(\alpha)}{\partial b}=\int_0^\alpha \frac{a^2\sin^2 t+b^2 \cos^2 t}{\sqrt{a^2\sin^2 t+b^2 \cos^2 t}}\,dt=\int_0^\alpha\sqrt{a^2\sin^2 t+b^2 \cos^2 t}\,dt $$
$$a\frac {\partial L}{\partial a}+b\frac {\partial L}{\partial b}= L \tag 1 $$
Could you please help me how to solve the partial differential equation?
Thanks a lot
Some boundary conditions: $$L(\alpha)|_{a=0}=\int_0^\alpha\sqrt{b^2 \cos^2 t}\,dt=\int_0^\alpha b \cos t\,dt =b \sin \alpha $$
$$L(\alpha)|_{b=0}=\int_0^\alpha\sqrt{a^2 \sin^2 t}\,dt=\int_0^\alpha a \sin t\,dt =-a (\cos \alpha- 1)= a (1- \cos \alpha) $$
$$L(\alpha)|_{b=a}=\int_0^\alpha\sqrt{a^2\sin^2 t+a^2 \cos^2 t}\,dt=\alpha a $$