Integral$ \int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$

Question

$$\int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$$

I tried to substitute $1+\cos^2x$ , tried to change cos in sin by complementary formula etc . But nothing seems to work out. I think it's not as simple as I thought ?

Answer 1

$$ I = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1+\cos^2 x}} dx = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{2-\sin^2 x}} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx $$

Recall the elliptic integral of the first and second kind, respectively:

$$F(\phi,k) = \int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2 \theta}}d\theta$$ $$E(\phi,k) = \int_{0}^{\phi} \sqrt{1-k^2\sin^2 \theta}d\theta$$

Is easy to show that

$$\int_{0}^{\phi} \frac{\sin^2 \theta}{ \sqrt{1-k^2\sin^2 \theta}} d\theta = \frac{F(\phi,k)-E(\phi,k)}{k^2}$$

$$\frac{F(\phi,k)-E(\phi,k)}{k^2} = \frac{1}{k^2}\int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2\phi}} - \sqrt{1-k^2\sin^2\phi} d\phi= \frac{1}{k^2}\int_{0}^{\phi} \left[\frac{1}{\sqrt{1-k^2\sin^2\phi}} -\frac{1-k^2\sin^2 \phi}{\sqrt{1-k^2\sin^2\phi}}\right]d\phi = \int_{0}^{\phi} \frac{\sin^2 \theta}{ \sqrt{1-k^2\sin^2 \theta}} d\theta$$

Hence:

$$ \int_{0}^{\phi} \frac{\cos^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \int_{0}^{\phi} \frac{1-\sin^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta - \int_{0}^{\phi} \frac{\sin^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = F(\phi,k) -\frac{F(\phi,k)-E(\phi,k)}{k^2} = \frac{E(\phi,k)-(1-k^2)F(\phi,k)}{k^2} = \frac{E(\phi,k)-k'^2F(\phi,k)}{k^2}$$ where $k$ is the modulus and $k'$ is the complementary modulus and $|k|<1, \quad k=\sqrt{1-k'^2}$.

Hence $$\boxed{\int_{0}^{\phi} \frac{\cos^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \frac{E(\phi,k)-k'^2F(\phi,k)}{k^2}}$$

Therefore

$$\boxed{I = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx = \frac{2}{\sqrt{2}}\left[E\left(\frac{\pi}{3},\frac{1}{\sqrt{2}}\right)-\frac{1}{2}F\left(\frac{\pi}{3},\frac{1}{\sqrt{2}}\right)\right]}$$

We have to take care evaluating elliptic functions since the software sometimes may use a slight different notation

Answer 2

Too long for a comment.

Depending on your level of knowledge about special functions, this problem is either very simple or very difficult.

For sure, in the second case, you could use numerical integration which, again, is simple.

There is another solution based on function approximation. Better than Taylor series, you can easy build the $[n,n]$ Padé approximant (because the first term of the Taylor expansion of the integrand is a constant). This is easy to do starting from the usual Taylor series.

Using $n=6$ we have $$\frac{\cos^2(x)}{\sqrt{1+\cos^2(x)}}=\frac 1 {\sqrt 2} \quad\frac {a+b x^2+c x^4+d x^6}{1+e x^2+f x^4+g x^6}$$ which is equivalent to the Taylor series to $O\left(x^{14}\right)$. Now, we can rewrite the denominator as $$1+e x^2+f x^4+g x^6=(x^2-r)(x^2-s)(x^2-t)$$ Using partial fraction decomposition $$\frac {a+b x^2+c x^4+d x^6}{(x^2-r)(x^2-s)(x^2-t)}=d+\frac R{x^2-r}+\frac S{x^2-s}+\frac T{x^2-t}$$ where $$R=\frac{a+b r+c r^2+d r^3}{(r-s) (r-t)}\quad S=\frac{a+b s+c s^2+d s^3}{(s-r) (s-t)}\quad T=\frac{a+b t+c t^2+d t^3}{(t-r) (t-s)}$$ This means that you face three simple antiderivatives and integrals.

I shall not give here the values of $(a,b,c,d,e,f)$ (this does not present much interest) but the final result of the approximation (at least its decimal representation) : $\color{red}{0.55682}36$ while the exact value is $\color{red}{0.5568281}$. Notice that using the Taylor series to $O\left(x^{14}\right)$, we would have obtained $\color{red}{0.5568}363$.

We could have done better making these approximations around $x=\frac \pi 6$. The result would have been $\color{red}{0.556828104}71$ to be compared to $\color{red}{0.55682810446}$.