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How to evaluate:
$$\int \sqrt{1+\cos^2x} dx$$

Is a simple antiderivative known to exist?

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Non-elementary, but traditional, in the family of elliptic integrals –  André Nicolas Aug 30 '12 at 14:31
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2 Answers 2

up vote 6 down vote accepted

A simple antiderivative is given by $$\tag{1} \int_0^x\sqrt{1+\cos^2t}\,dt. $$ This is an example of what is called an elliptic integral and it cannot be expressed in terms of other well-known elementary functions (which is what you probably mean by "simple", and doesn't agree with what I would call "simple").

But let me stress my point: the expression in (1) can be used to approximate values of the function, and in that sense it is not necessarily worse than, say $\log x$.

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This question is much like the other one.

Using $\cos^2(x) = 1-\sin^2(x)$ $$ \int \sqrt{1+\cos^2(x)} \mathrm{d} x = \int \sqrt{2-\sin^2(x)} \mathrm{d} x = \sqrt{2} \int \sqrt{1-\frac{1}{2} \sin^2(x)} \mathrm{d} x = \sqrt{2} \operatorname{E}\left(x, \frac{1}{2}\right) + C $$ where $\mathrm{E}(x, m)$ denotes the incomplete elliptic integral of the second kind: $$ \operatorname{E}\left(\phi, m\right) = \int_0^\phi \sqrt{1-m \sin^2(\varphi)} \mathrm{d} \varphi $$ In particular, no elementary andi-derivative exists.

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