Is it still pretty?

Consider this integral

$\displaystyle \int_{0}^{\pi/2}\frac{1}{1+\tan^{3}(x)} dx = \int_{0}^{\pi/4}\frac{1}{1+\tan^{3}(x)} dx + \int_{\pi/4}^{\pi/2}\frac{1}{1+\tan^{3}(x)} dx$

then in the second term we change from x to u such that $\displaystyle u=x-\pi/2$

$\displaystyle \int_{-\pi/4}^{0}\frac{\tan^{3}(x)}{\tan^{3}(x)-1} dx$ then change into $u'=-u$

$\displaystyle \int_{0}^{\pi/4}\frac{\tan^{3}(x)}{1+\tan^{3}(x)} dx$

combine these two term togather.

$\displaystyle \int_{0}^{\pi/4}(\frac{1}{1+\tan^{3}(x)}+\frac{\tan^{3}(x)}{1+\tan^{3}(x)}) dx$

$\displaystyle \int_{0}^{\pi/4} dx=\frac{\pi}{4}$