<s>[INST] How to integrate $\int\limits_{0}^{\pi/2}\frac{dx}{\cos^3{x}+\sin^3{x}}$? I have$$\int\limits_{0}^{\pi/2}\frac{\text{d}x}{\cos^3{x}+\sin^3{x}}$$
Tangent half-angle substitution gives a fourth-degree polynomial in the denominator that is difficult to factor.
 [/INST] $$I = \int_{0}^{\pi/2}\frac{dx}{\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\left(1-\frac{1}{2}\sin(2x)\right)}=\int_{-\pi/4}^{\pi/4}\frac{dx}{\sqrt{2}\cos(x)\left(1-\frac{1}{2}\cos(2x)\right)} $$
hence, through the substitution $x=\arcsin t$:
$$ I = \sqrt{2}\int_{0}^{\pi/4}\frac{dx}{\cos(x)\left(\frac{3}{2}-\cos^2 x\right)}=\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{dt}{(1-t^2)\left(\frac{1}{2}+t^2\right)}$$
and the last integral is perfectly manageable through partial fraction decomposition.
The outcome is:

$$ \int_{0}^{\pi/2}\frac{d\theta}{\sin^3\theta+\cos^3\theta} = \color{red}{\frac{\pi}{3}+\frac{2\sqrt{2}}{3}\,\log\left(1+\sqrt{2}\right)}.$$

</s>