Integral of $\sqrt{c-\cos(x)}$ where c is a constant

Question

I am having trouble calculating the following integral:

$$ {\int \sqrt{c-\cos(t)} dt} $$

where $c$ is a constant.

In the very specific case where $c=1$, it is rather easy, but I cannot seem to be able to generalize that.

I tried running it with Wolfram but it produces higher-than-my-paygrade math (example) about elliptic integrals and I can't help but think it should be easier than that.

Answer 1

No, it is not easier than that. Wolfram showing you "weird" functions is a sure sign, if not a proof*.

The elliptic integrals have become very classical. https://en.wikipedia.org/wiki/Elliptic_integral

It is easy to understand why the case of $c=1$ is simpler. In the plots you see that the blue curve is a (rectified) sinusoid, while the others are not, as can be checked analytically.

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*A proof is possible but highly technical, based on Liouville's theorem. https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)

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Also note that a simple integrand rarely means a simple antiderivative. Try integrate 1/(x^5+1).

Answer 2

For the specific case where $~=1~$, this integral is rather easy because here you can use a trigonometric formula $~1-\cos t~=~2\sin^2\left(\frac{t}{2}\right)~$ directly in the second line and got a known trigonometric function $\big($here it is $\sin\left(\frac{t}{2}\right)\big)$ which can be integrable by known formula (as we have the knowledge that anti-derivative of $~\cos t~$ is $~\sin t~+~c$, where $c$ is a constant) . But for the value $~c=2~$or more you can't find such form.

For example, if you take $~c=2~$, you got (after some steps) a function as $~\sqrt{2\sin^2\left(\dfrac{t}{2}\right)+1}~$. There is no elementary function whose derivative is $~\sqrt{2\sin^2\left(\dfrac{t}{2}\right)+1}~$. Here you have to allow yourself to know about a special integral called incomplete elliptic integral of the second kind.

• Here is the complete solution of the given integral.

\begin{equation} I={\displaystyle\int}\sqrt{c-\cos\left(t\right)}\,\mathrm{d}t\\ ={\displaystyle\int}\sqrt{c-1}\cdot\sqrt{\dfrac{2\sin^2\left(\frac{t}{2}\right)}{c-1}+1}\,~\mathrm{d}t\\ \end{equation} Substituting $~u=\dfrac{t}{2}~$, so $~\mathrm{d}t=2\,\mathrm{d}u~$. Now
\begin{equation} I=\class{steps-node}{\cssId{steps-node-1}{2\sqrt{c-1}}}{\cdot}{\displaystyle\int}\sqrt{\dfrac{2\sin^2\left(u\right)}{c-1}+1}\,~\mathrm{d}u\tag1 \end{equation}

We know that \begin{equation} {\displaystyle\int}\sqrt{\dfrac{2\sin^2\left(u\right)}{c-1}+1}\,\mathrm{d}u~~ =~~\operatorname{E}\left(u\,\middle|\,-\dfrac{2}{c-1}\right) \end{equation} This is a special integral (incomplete elliptic integral of the second kind).

Hence from $(1)$ , \begin{equation} I~=~2\sqrt{c-1}~\operatorname{E}\left(u\,\middle|\,-\dfrac{2}{c-1}\right)~+~K\\ =2\sqrt{c-1}\operatorname{E}\left(\dfrac{t}{2}\,\middle|\,-\dfrac{2}{c-1}\right)~+~K \end{equation} where $~K~$ is constant and $~c\ne 1~$.

For $~c=1~$, the value of the integral is $$~I~=~-2^\frac{3}{2}\cos\left(\dfrac{t}{2}\right)+K~$$

Answer 3

$$=\int{\sqrt {c -1}\sqrt{\frac{2sin^2\frac{x}{2}}{c-1}+1dt}}$$ substitute $u=\frac{t}{2}$ -> $\frac{du}{dt}=\frac{1}{2}$ -> $dt = 2du$ $$= 2\sqrt{c-1}.\int \sqrt{\frac{2sin^2u}{c-1}+1}du$$ If you solve this integral you will find the solution as $$2\sqrt{c-1}E(\frac{x}{2}|-\frac{2}{c-1})+C$$