$\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.

Solution 1:

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 $\pi/4$.

Solution 2:

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.

$$I=\int_{-\pi/2}^{\pi/2}\dfrac{1}{1+\alpha^x}\underbrace{\biggl(\dfrac{\sin^{2012}x}{\sin^{2012}x+\cos^{2012}x}\biggr)}_{\text{even function}}\mathrm dx = \int_{-\pi/2}^{\pi/2}\dfrac{\alpha^x}{1+\alpha^x}\biggl(\dfrac{\sin^{2012}x}{\sin^{2012}x+\cos^{2012}x}\biggr)\mathrm dx\\\implies 2I=\int_{-\pi/2}^{\pi/2}\dfrac{\mathrm dx}{1+\tan^{2012}x}=2\int_{0}^{\pi/2}\dfrac{\mathrm dx}{1+\tan^{2012}x}\implies I=\dfrac{\pi}{4}$$

Solution 3:

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 $\alpha>0$.