Evaluation of a definite integral with square root in denominator

Question

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)

Answer 1

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

Answer 2

Hint. It comes from $$ |\alpha|<x \implies\alpha^2-x^2 <0 \quad \text{giving}\quad \sqrt{\alpha^2-x^2}\in \mathbb{R} i. $$ One has, for $0<\alpha<\pi$, $$ \int_0^{\pi } \frac{x \cos (x)}{\sqrt{\alpha^2-x^2}} \, dx=\underbrace{\int_0^{\alpha} \frac{x \cos (x)}{\sqrt{\alpha^2-x^2}} \, dx}_{\large \,\in\, \mathbb{R}}+\underbrace{\int_{\alpha}^{\pi } \frac{x \cos (x)}{\sqrt{\alpha^2-x^2}} \, dx}_{\large \,\in\, \mathbb{R} i}. $$