Definite integral of a positive continuous function equals zero? Let's calculate $$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x}$$
We have
$$\int \frac {dx}{\sin^6x + \cos^6x} = \int \frac {dx}{1 - \frac 34 \sin^2{2x}}$$
now we substitute $u = \tan 2x$, and get
$$\int \frac {dx}{1 - \frac 34 \sin^2{2x}} = \frac{1}{2} \int \frac {du}{1 + \frac 14 u^2} = \tan^{-1}\frac u2 + C= \tan^{-1}\left(\frac12 \tan 2x\right) + C = F(x)$$
Now, evaluating the primitive function at $x = 0, x = \frac{\pi}2$, we get
$$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x} = F\left(\frac {\pi}2\right) - F(0) = 0 - 0 = 0$$
But the integrand is positive and continuous, so the integral should be positive!! What have I done wrong?

One thing I can see: in your substitution $\;u=\tan 2x\;$ on $\;[0,\pi/2]\;$ , you get 
$$0\le x\le\frac\pi2\implies 0\le 2x\le\pi\implies u=\tan2x$$
is not defined on $\;[0,\pi]\;$ , which renders the substitution incorrect.
