Evaluation of a definite integral with square root in denominator The definite integral
$\displaystyle \int_0^{\pi } \frac{x \cos (x)}{\sqrt{\alpha^2-x^2}} \, dx$
is evaluated numerically.
After numerical evaluation with a CAS, it is found that the integral has real numerical values for $\alpha \geq \pi$, and complex numerical values for $\alpha<\pi$.
How is this proven or explained?
Does it hold that for any definite integral
$\displaystyle \int_0^{\beta } \frac{x \cos (x)}{\sqrt{\alpha^2-x^2}} \, dx$
the numerical values are real for $\alpha \geq \beta$? ($\alpha$ and $\beta$ are real)

If $0<\alpha < \beta$, then $a^2-x^2<0$ for $\alpha < x < \beta$, so you're taking the square root of negative numbers.
