I need to calculate the area between $y=0$ and $y=\sqrt{\cos(x)}$ between $x=\frac{\pi}{4}$ $x=\frac{\pi}{2}$. I tried integrating $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{\cos(x)}dx$ but it I don't know how. I tried letting $r=tg(x/2)$ but seems that doesn't help. Also I tried changing it to $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{\cos(x)}\frac{\sin(x)}{\sin(x)}dx$ and integrate by substitiution but that didn't work either. How can I integrate this expression?
[Math] Integrating $\sqrt{\cos(x)}$
calculusintegration
Best Answer
This is going to be an elliptic integral of some sort. Write
$$\int_{\pi/4}^{\pi/2} dx \: \sqrt{\cos{x}} = \int_{0}^{\pi/2} dx \: \sqrt{\cos{x}} - \int_{0}^{\pi/4} dx \: \sqrt{\cos{x}}$$
In the first integral, let $u=\sin{x}$, $dx=du (1-u^2)^{-1/2}$; the integral becomes
$$\int_{0}^1 du \: (1-u^2)^{-1/4} = \underbrace{\frac12 \int_{0}^1 dv\: v^{-1/2} (1-v)^{-1/4}}_{v=u^2} = \frac12 \frac{\Gamma\left(\frac12 \right) \Gamma\left(\frac34 \right)}{\Gamma\left(\frac54 \right)} = \sqrt{\frac{2}{\pi}}\Gamma\left(\frac34 \right)^2 $$
In the second integral, rewrite as
$$\underbrace{\int_0^{\pi/4} dx \: \left(1-2 \sin^2{\frac{x}{2}}\right)^{1/2}}_{\text{half-angle formula}} = 2 \int_0^{\pi/8} du \: \left(1-2 \sin^2{u}\right)^{1/2} = 2 E\left(\frac{\pi}{8}\vert 2\right)$$
where I use the Wolfram definition of the elliptic integral
$$E(\phi \vert m) = \int_0^{\phi} du \: \left(1-m \,\sin^2{u}\right)^{1/2}$$
Therefore
$$\int_{\pi/4}^{\pi/2} dx \: \sqrt{\cos{x}} = \sqrt{\frac{2}{\pi}}\Gamma\left(\frac34 \right)^2 - 2 E\left(\frac{\pi}{8}\vert 2\right) \approx 1.19814-2 (0.372152) = 0.452837$$
I doubt you will improve much on that.