Two integrals, each easier than it looks

I start with an integral that stymied a bunch of people for the better part of an hour. The following solution had a lot of people slapping their heads.

The problem is to evaluate, for any real $\alpha$, the following integral:

$$\int_0^{\pi/2} \frac{dx}{1+\tan^{\alpha}{x}}$$

Solution: use the fact that

$$\tan{\left (\frac{\pi}{2}-x\right)} = \frac{1}{\tan{x}}$$

i.e.,

$$\frac1{1+\tan^{\alpha}{x}} = 1-\frac{\tan^{\alpha}{x}}{1+\tan^{\alpha}{x}} = 1-\frac1{1+\frac1{\tan^{\alpha}{x}}} = 1-\frac1{1+\tan^{\alpha}{\left (\frac{\pi}{2}-x\right)}}$$

Therefore, if the sought-after integral is $I$, then

$$I = \frac{\pi}{2}-I \implies I=\frac{\pi}{4}$$

The second problem is to evaluate, for $\alpha \gt 0$:

$$\displaystyle \int_{-\pi/2}^{\pi/2} dx\, \frac{\sin^{2012}{x}}{\left(1+ \alpha^x\right)\left(\sin^{2012} {x}+\cos^{2012}{x}\right)}$$

Rewrite as

$$\int_{-\pi/2}^{\pi/2} \frac{dx}{1+e^{b x}} \frac1{1+\cot^{2012}{x}}$$

where $b=\log{\alpha}$. This integral is equal to

$$\underbrace{\int_{-\pi/2}^0 \frac{dx}{1+e^{b x}} \frac1{1+\cot^{2012}{x}}}_{x \mapsto -x} + \int_0^{\pi/2} \frac{dx}{1+e^{b x}} \frac1{1+\cot^{2012}{x}}$$

or

$$\int_0^{\pi/2} \frac{dx}{1+\cot^{2012}{x}}\underbrace{\left (\frac1{1+e^{b x}} + \frac1{1+e^{-b x}} \right )}_{\text{this is} = 1} = \int_0^{\pi/2} \frac{dx}{1+\cot^{2012}{x}}$$

or

$$\int_0^{\pi/2} \frac{dx}{1+\tan^{2012}{x}}$$

which, from the first result above is $\pi/4$.