Why can't I substitute $u=ie^{-it}$ on integral $\int_{-\frac{\pi}{2}}^\frac{\pi}{2} (e^{-2it}+1)^x\ln(e^{2it}+1)dt$?

Question

I wanted to use substitution $u=ie^{-it}$ to simplify following integral: $$\int_{-\frac{\pi}{2}}^\frac{\pi}{2} (e^{-2it}+1)^x\ln(e^{2it}+1)dt$$ After substituting I got following integral: $$i\int_{-1}^1 \frac{(1-u^2)^x\ln(1-\frac{1}{u^2})}{u}du$$ This is the integral of the odd function over symmetric interval and therefore it is equal to $0$. However this is not proper result (For example when $x=1$ the integral should be equal to $\pi$).
What did I do wrong?

Answer 1

In the first integral, when $t$ is from $-\frac{\pi}{2}$ via $0$ to $\frac{\pi}{2}$, $ie^{-2it}$ is from $-1$ via $i$ to $1$. However $u$ is from $-1$ via $0$ to $1$ in the second integral. They go through different paths. They coincide only under certain conditions.