Can this be correct $I_{\alpha}=\int_{0}^{\pi/2}\mathrm dx {\sin(2x)\over 1+\tan^{\alpha}x}={1\over 2}?$ $$I_{\alpha}=\int_{0}^{\pi/2}\mathrm dx {\sin(2x)\over 1+\tan^{\alpha}x}={1\over 2}\tag1$$
$\alpha \ge 0$
The indefinite integral for $\alpha=1,2,3...$
$$I_1={1\over 4}\sin(2x)-{1\over 4}\cos(2x)-{1\over 2}\log[\sin x+\cos x]+C$$
$$I_2=-{1\over 2}\cos^4(x)+C$$
$$I_3=-{1\over 4}\sin(2x)+{1\over 4}\cos(2x)-{2\over 3}\log[2-\cos 2x]-{1\over 6}\ln[\sin x +\cos x]+C$$
The indefinite integral looks complicate for $\alpha$, how can $(1)$ gives ${1\over 2}?$
 A: Hint:
As $\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$
if  $\displaystyle I=\int_a^bf(x)dx$
$$I+I=\int_a^b\{f(x)dx+f(a+b-x)\}dx=?$$
A: By differentiating $I_\alpha$ with respect to $\alpha$, one obtains:
$$I’=\int^{\pi /2}_0 \sin(2x)\frac{\ln(\tan(x))\tan^\alpha (x)}{(1+\tan^{\alpha}(x))^2}dx$$
By the substitution $x \to \pi/2-x$, one obtains:
$$I’=\int^{\pi /2}_0 \sin(2x)\frac{\ln(\cot(x))\cot^\alpha (x)}{(1+\cot^{\alpha}(x))^2}dx=-\int^{\pi /2}_0 \sin(2x)\frac{\ln(\tan(x))\tan^\alpha (x)}{(1+\tan^{\alpha}(x))^2}dx=-I’$$
Since $I’=-I’$, $I’=0$.
Thus, no matter what value of $\alpha$ you substitute in, the integral is still the same.
Substitute in $\alpha=0$, you will obtain $I=\frac12$ easily.
What is even more interesting is the case $\alpha =\infty$. For $x<\pi/4$, $\tan(x)<0$, $\tan^\infty (x)=0$, thus the integrand reduces to $\sin(2x)$; for $x>\pi/4$, $\tan^\infty (x)=\infty$, the integrand is zero. So, 
$$I_\infty=\int^{\pi/4}_0 \sin(2x)dx=\frac12$$
Similarly,
$$I_{-\infty}=\int^{\pi/2}_{\pi/4}\sin(2x)dx=\frac12$$
A: As @labbhattacharjee suggested, use the property   
$\int_{a}^{b}f(x)=\int_{a}^{b}f(a+b-x)$    
$\implies I=\int_{0}^{\pi/2}\frac{\sin 2x}{1+\tan^\alpha x}dx=\int_{0}^{\pi/2}\frac{\sin 2x}{1+\cot^\alpha x}dx$    
$\implies 2I=\int_{0}^{\pi/2}\frac{\sin 2x}{1+\tan^\alpha x}+\frac{\sin 2x}{1+\cot^\alpha x}dx$    
$\implies 2I= \int_{0}^{\pi/2}(\frac{\sin 2x(1+\tan^\alpha x)}{1+\tan^\alpha x}+\frac{\sin 2x(1+\cot^\alpha x)}{1+\cot^\alpha x})dx$    
$\implies 2I=\int_{0}^{\pi/2}(\frac{\sin 2x(1+\tan^\alpha x+\cot^\alpha x +1)}{1+\tan^\alpha x+\cot^\alpha x +1})dx $    
$\implies 2I =\int_{0}^{\pi/2}(\sin 2x) dx $    
$\therefore  2I=1$    
$\therefore I=\frac{1}{2}$
