Is it possible to compute $\int_0^{\pi/2}\sqrt{a^2 \sin^2 x + b^2 \cos^2 x}\ dx$.

Question

Basically, this integral, except the $\ln$ is replaced with $\sqrt\cdot$.

In a question we were asked to find the circumference of part of an ellipse, and we got this integral, so I was wondering if we can explicitly compute it.

Answer 1

As stated in the comments, this integral is an example of an Elliptic integral (named this way because, as you said in your question, they're related to the perimeter of ellipses).

In general, these integrals are non-elementary, so to get a closed-form answer we need to define a new special function. Specifically, we'll use a special function called the "Elliptic integral of the Second Kind", which is defined as $$ E(\color{blue}{\phi}, \color{purple}{k}) = \int_0^{\color{blue}{\phi}}\sqrt{1-\color{purple}{k}^2\sin^2(x)} \, \mathrm{d}x $$ Using the function defined above we can evaluate the integral as follows: \begin{align} \int_0^{\frac{\pi}{2}}\sqrt{a^2 \sin^2(x) + b^2 \cos^2(x)}\, \mathrm{d}x & =\int_0^{\frac{\pi}{2}}\sqrt{a^2 \sin^2(x) + b^2\left[1 - \sin^2(x) \right]}\, \mathrm{d}x\\ & =\int_0^{\frac{\pi}{2}}\sqrt{b^2 -(b^2 -a^2) \sin^2(x) }\, \mathrm{d}x\\ & =|b|\int_0^{\color{blue}{\frac{\pi}{2}}}\sqrt{ 1-\left(\color{purple}{\sqrt{1 -\frac{a^2}{b^2}}}\right)^2 \sin^2(x) }\, \mathrm{d}x\\ & = \boxed{|b|E \left(\frac{\pi}{2} , \sqrt{1 - \frac{a^2}{b^2}}\right)} \end{align} Hope this helps!

Answer 2

notice: $$b^2\cos^2x+a^2\sin^2x\\ =b^2\cos^2x+b^2\sin^2x+a^2\sin^2x-b^2\sin^2x\\ =b^2+(a^2-b^2)\sin^2x\\ =b^2\left[1+\underbrace{\left(1-\left(\frac ab\right)^2\right)}_{k^2}\sin^2x\right]\\ =b^2(1-k^2\sin^2x)$$ and so your integral can be written as: $$|b|\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx$$

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Now the definition of the complete elliptic integral of the second kind is: $$E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx$$ which is exactly what we have. You can therefore say: $$I=|b|E\left(\sqrt{1-\left(\frac ab\right)^2}\right)$$ which is unfortunately the nicest form you will get this into for any $a,b$.

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It is worth noting that there are 3 elliptic integrals (first, second and third kind) and you have the complete form here due to the limits. These integrals also exist in incomplete form such as the incomplete elliptic integral of the second kind where the upper limit is open ended and is denoted $E(\varphi,k)$