<s>[INST] Convergence of integral with trigonometry 
*

*$\displaystyle \int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}} dx$



Obviously, problem in $ -\frac{\pi}{4} $ as $ \lim_{x\to -\frac{\pi}{4}^+} = \infty $, integral is positive on whole segment, but I can't use any usual rules to prove it's convergence or simly divide it inside the root. I guess, i have to bound it with converging function, but can't find a good one aswell. Any thoughts?
 [/INST] Recall that


*

*$\cos x - \sin x=\sqrt 2 \sin \left(\frac{\pi}4-x\right)$

*$\cos x + \sin x=\sqrt 2 \sin \left(\frac{\pi}4+x\right)$
then
$$ =\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)^{\frac{1}{3}} dx
=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\sin \left(\frac{\pi}4-x\right)}{\sin \left(\frac{\pi}4+x\right)}\right)^{\frac{1}{3}} dx$$
Assume $y=x+\frac{\pi}4$ then we have
$$\int_{-\frac{\pi}{4}}^\frac{\pi}{4} \left(\frac{\sin \left(\frac{\pi}4-x\right)}{\sin \left(\frac{\pi}4+x\right)}\right)^{\frac{1}{3}} dx
=\int_{0}^\frac{\pi}{2} \left(\frac{\sin \left(\frac{\pi}2-y\right)}{\sin y}\right)^{\frac{1}{3}} dy
=\int_{0}^\frac{\pi}{2} \left(\frac{1}{\tan y}\right)^{\frac{1}{3}} dy$$
and
$$\left(\frac{1}{\tan y}\right)^{\frac{1}{3}}\sim\frac1{\sqrt[3] y} $$
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