I am looking into the proof of Ramanujan's approximation formula for the partition function $p(n)$ by Stein and Shakrachi. I am confused about one step towards the end of the proof just before we apply the method of steepest descent. Stein obtains
$$-\mu^{\frac{3}{2}}\int_{0}^{\pi}e^{2s\sin{(\theta)}}e^{\frac{3i\theta}{2}}\sqrt{i} \, d\theta = \mu^{\frac{3}{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{2s\cos{(\theta)}}\left(\cos{\left(\frac{3\theta}{2}\right)}+i\sin{\left(\frac{3\theta}{2}\right)}\right) \, d\theta.$$
How we go from one integral to the other is what is confusing me at the moment. Is there a transformation identity? Or am I missing something elementary?
First we can replace $\theta $ with $\pi-\theta$ and get the integrand as $$\exp\left(2s\sin \theta+\frac{i\pi} {4}+\frac{3i\pi}{2}-\frac{3i\theta}{2}\right)$$ Next we put $\theta=\pi/2-t$ to get the integrand as $$\exp\left(2s\cos t+i\pi+\frac{3it}{2}\right)$$ The $i\pi$ leads to a negative sign, another negative comes from $-dt$ and finally one more negative sign comes from the reversal of integral limits $[\pi/2,-\pi/2]$ to $[-\pi/2,\pi/2]$. So all in all we have the change seen as a "-" sign.