$\int_{-\pi/2}^{\pi/2} \frac{\sin^{2012}{x}}{\left(1+ \alpha^x\right)\left(\sin^{2012} {x}+\cos^{2012}{x}\right)}\;{dx} $ For $\alpha\in\mathbb{R^+}$, evaluate
$$\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^{2012}{x}}{\left(1+ \alpha^x\right)\left(\sin^{2012}  {x}+\cos^{2012}{x}\right)}\;{dx} $$
Can I have a hint on this? I tried substituting $x=-t$ but got nowhere.
 A: This can be solved simply by using reflection. In fact the substitution you made namely $x\to -t$ shall lead you to the solution. You simply need to use the reflection property twice.
$$I=\int_{-\pi/2}^{\pi/2}\dfrac{1}{1+\alpha^x}\underbrace{\biggl(\dfrac{\sin^{2012}x}{\sin^{2012}x+\cos^{2012}x}\biggr)}_{\text{even function}}\mathrm dx = \int_{-\pi/2}^{\pi/2}\dfrac{\alpha^x}{1+\alpha^x}\biggl(\dfrac{\sin^{2012}x}{\sin^{2012}x+\cos^{2012}x}\biggr)\mathrm dx\\\implies 2I=\int_{-\pi/2}^{\pi/2}\dfrac{\mathrm dx}{1+\tan^{2012}x}=2\int_{0}^{\pi/2}\dfrac{\mathrm dx}{1+\tan^{2012}x}\implies I=\dfrac{\pi}{4}$$
A: 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 by this result, is $\pi/4$.
A: A non exact solution (but very very close) can be obtained by looking at:
$$g(x)=\frac{\sin^{2012} x}{\sin^{2012} x+ \cos^{2012} x}$$
For all practical purposes, $g(x)$ may be approximated by:
$$g(x) =
\left\{
 \begin{array}{ll}
  1  & \mbox{if } |x| > \pi/4 \\
  0 & \mbox{if } |x| < \pi/4
 \end{array}
\right.$$
This means your integral gets boiled down to:
$$\int\frac{dx}{1+\alpha^x} = x - \frac{\log(1 + \alpha^x)}{\log \alpha} $$
On the new intervals which is equal to: $$\pi/4$$ for every value $\alpha>0$.
