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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: $$L(\alpha)|_{a=0}=\int_0^\alpha\sqrt{b^2 \cos^2 t}\,dt=\int_0^\alpha b \cos t\,dt =b \sin \alpha $$

$$L(\alpha)|_{b=0}=\int_0^\alpha\sqrt{a^2 \sin^2 t}\,dt=\int_0^\alpha a \sin t\,dt =-a (\cos \alpha- 1)= a (1- \cos \alpha) $$

$$L(\alpha)|_{b=a}=\int_0^\alpha\sqrt{a^2\sin^2 t+a^2 \cos^2 t}\,dt=\alpha a $$

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    Hint: http://en.wikipedia.org/wiki/Method_of_characteristics2012-09-13
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    @Mathlover, although you create the PDE correctly, in fact express $\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2 t}~dt$ in series expression should only be done by other means method and the PDE approach is no helpful. You are an experienced user, why you ask such wrongly-directed question?2012-09-14
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    For the PDE aspect, see [this thread](http://math.stackexchange.com/questions/173532/coordinate-method-for-solving-first-order-linear-pde). But you should not try to reinvent the wheel. There are [hundreds of formulas](http://functions.wolfram.com/EllipticIntegrals/) for elliptic integrals, including dozens of series representations.2012-09-14
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    @doraemonpaul: I have been self-studying on elliptic integrals. I like to try different approach to a problem. I am not so much experienced about some subjects such as PDEs. Actually I knew binom method but I tried to find a solution via a different way . Maybe I have lost myself in big PDE Ocean. Thanks for answer.2012-09-14
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    @LVK: Thanks for advice. You are right but to rediscover some relations gives me deep understanding. It is just self-learning method. I am trying to learn to catch fish not to wait for a ready one in my stomach. Thank you for your understanding and links. I have been checking them.2012-09-14

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