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?
 A: The problem is indeed in $-\pi/4$. 
The Taylor expansion around $-\pi/4$ for the sine:
\begin{align}
\sin(x) = -\frac{1}{2}\sqrt[]{2}+\frac{1}{2}\sqrt[]{2}  (x+\pi/4) + O((x+\pi/4)^2)
\end{align}
and for the cosine:
\begin{align}
\cos(x) = \frac{1}{2}\sqrt[]{2}+\frac{1}{2}\sqrt[]{2}  (x+\pi/4)+O((x+\pi/4)^2)
\end{align}
Therefore:
\begin{align}
\cos(x)+\sin(x) = \sqrt[]{2}(x+\pi/4) + O((x+\pi/4)^2)
\end{align}
So,  near $-\pi/4$,  the integrand  behaves as: $$\frac{1}{(\sqrt[]{2}(x+\pi/4))^{1/3}}$$ That means that the original integral converges. 
There are some details left for you, for example finding the convergent integral that dominates your integral, to make this argument fully rigorous.
A: $$I=\int_{-\frac{\pi}{4}}^\frac{\pi}{4} (\frac{\cos x - \sin x}{\cos x + \sin x})^{\frac{1}{3}}, dx$$
$1)sin(x)+cos(x)=Rcos(x-a)\therefore R=\sqrt{2}cos(x-\frac{\pi}{4})$
$$sin(x)+cos(x)=\sqrt{2}cos(x-\frac{\pi}{4})$$
do the same for $cos(x)-sin(x)$, so
$$cos(x)-sin(x)=\sqrt{2}cos(x+\frac{\pi}{4})$$
$$\therefore\frac{\cos x - \sin x}{\cos x + \sin x}=\frac{\sqrt{2}cos(x+\frac{\pi}{4})}{\sqrt{2}cos(x-\frac{\pi}{4})}=\frac{cos(x+\frac{\pi}{4})}{cos(x-\frac{\pi}{4})}$$
Can you do anything with this?
A: Starting from gimusi's answer, consider
$$I=\int \frac{dy}{\tan^{1/3}(y)} $$
Let 
$$y=\tan ^{-1}\left(z^3\right)\implies dy=\frac{3 z^2}{z^6+1}\,dz$$ which make
$$I=\int \frac{3 z}{z^6+1}\,dz=\int \frac{3 z}{(z^2+1)(z^4-z^2+1)}\,dz$$ Contiue the factorization of $z^4-z^2+1$.
I am sure that you can take it from here.
A: 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
$$\left(\frac{1}{\tan y}\right)^{\frac{1}{3}}\sim\frac1{\sqrt[3] y} $$
