So, my problem is shown here 
.
In the second last integral we have :
$$\frac{M}{R^{k-1}} \int_{\theta_0}^{\pi/2} e^{(R\cos{\theta})t} \Bbb d\theta = \frac{M}{R^{k-1}} \int_{0}^{\phi_0} e^{(R\sin{\theta})t} \Bbb d\phi \tag{Eq. 1}$$
As we know from the screenshot, we set $$\theta = \pi/2 - \phi\tag{Eq. 2}$$. So the left hand side of Eq. $1$ should be:
$$\frac{M}{R^{k-1}} \int_{\pi/2 - \theta_0}^{0} e^{(R\sin{\phi})t} \Bbb d\phi\tag{Eq. 3}$$
Then, from Eq. $2$ we can substitute the lower bound of integral Eq. $3$ and it becomes:
$$\frac{M}{R^{k-1}} \int_{\phi_0}^{0} e^{(R\sin{\phi})t} \Bbb d\phi$$
But as we know in real integration if we want to change lower and upper bound we should multiply the integrand with $(-1)$. So:
$$\frac{M}{R^{k-1}} \int_{0}^{\phi_0} -e^{(R\sin{\phi})t} \Bbb d\phi$$
But wait! Let's see what's shown on the screenshot, it's clear that there's no negative term there. And I believe $M$, $R$, and $k$ have nothing to do with the integral since they're all positive. Or maybe the negative sign vanishes because of absolute value? But we've already removed the absolute value.
So, my question is Where did the negative sign go or maybe I missed something very obvious? Hope you can help me. I spent 2 hours to read this over and over again but still don't understand.
Thanks in advance!
At first glance, since $ \theta = \pi/2 - \phi $, you have an extra minus, since $ d\theta = - d\phi $. So the two minus signs cancel out.