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I haven't seen integration by parts written that way before, and the derivation you describe seems overly complicated, albeit not for the $I$ stuff. In particular, I don't understand the purpose of the trigonometric substitutions as the integral should be solvable already without them. Maybe there was a reason in context. I assume the book had some reason for using that derivation.
Anyways, I've usually seen integration by parts written $$\big\[uv\big\]\_{x=t\_0}^{x=t} = \int\_{t\_0}^t u dv + \int\_{t\_0}^t v du$$ and usually solved for one of the integrals, e.g. $$\int\_{t\_0}^t v du = \big\[uv\big\]\_{x=t\_0}^{x=t} - \int\_{t\_0}^t u dv$$ where $\big[uv\big]\_{x=t\_0}^{x=t}$ means evaluate $uv$ at $t$ and subtract $uv$ evaluated at $t_0$, i.e. $\big[f(x)\big]\_{x=t\_0}^{x=t} = f(t) - f(t\_0)$. (This operation can actually be understood as a $0$-dimensional contour integral.) I'll abbreviate this as $\big[f(x)]\_{t\_0}^t$.
The derivation of this is quite simple by integrating the product rule, i.e. we have $$\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$ or often written $$d(uv) = udv + vdu$$ Integrating and applying the fundamental theorem of calculus gives you integration by parts as I've written it above.
For your specific example, I (personally) would choose $du = e^{ax}dx$ and $v = \cos(bx + c)$. (The other choice is equally valid, though in many cases one choice is significantly easier than the other. Here, it doesn't make that much of a difference.) I'll also set $t\_0 = 0$ as it only affects the result by a constant term. This gives $$\int\_0^t e^{ax}\cos(bx + c)dx = \left[\frac{1}{a}e^{ax}\cos(bx + c)\right]\_0^t + \frac{b}{a}\int\_0^t e^{ax}\sin(bx+c)dx$$
We now apply integration by parts again to the second integral and again choosing $du = e^{ax}dx$ and $v = \sin(bx + c)$ producing $$\int\_0^t e^{ax}\sin(bx+c)dx = \left[\frac{1}{a}e^{ax}\sin(bx + c)\right]\_0^t - \frac{b}{a}\int\_0^t e^{ax}\cos(bx + c)dx$$ We see have reproduced the integral we were originally trying to solve. This is a common pattern and usually a good thing. Sometimes the integral might be one of the intermediate steps. The important point is once we're in this situation, we now have an "algebraic" equation. This is indeed made clearer by naming the integral $I$, i.e. setting $I=\int\_0^t e^{ax}\cos(bx + c)dx$. Substituting this second integration by parts back into the first and using this substitution, we get $$I = \left[\frac{1}{a}e^{ax}\cos(bx + c)\right]\_0^t + \frac{b}{a}\left(\left[\frac{1}{a}e^{ax}\sin(bx + c)\right]\_0^t - \frac{b}{a}I\right)$$
The important point here is that, as an equation in $I$, this no longer involves any integration. Distributing the $\frac{b}{a}$ and multiplying the whole equation by $a^2$ gives $$a^2I = \left[ae^{ax}\cos(bx + c) ight]\_0^t + b\left[e^{ax}\sin(bx + c) ight]\_0^t - b^2I$$ We can easily solve for $I$ (the original integral) by adding $b^2I$ to both sides then dividing by $a^2 + b^2$. $$I = \frac{\left[ae^{ax}\cos(bx + c) ight]\_0^t + b\left[e^{ax}\sin(bx + c) ight]\_0^t}{a^2 + b^2}$$ Simplifying a bit produces $$I = \frac{ae^{at}\cos(bt + c) - a\cos(c) + be^{at}\sin(bt + c)}{a^2 + b^2}$$ The correctness of this expression can be verified by differentiating with respect to $t$ and evaluating at $t=0$ which should result in $0$.
(Connecting to your other question, we could rewrite $\cos(bx + c)$ as $(e^{i(bx + c)} + e^{-i(bx + c)})/2$ and the integral would split into two straightforward integrals of exponential functions.)

I haven't seen integration by parts written that way before, and the derivation you describe seems overly complicated, albeit not for the $I$ stuff. In particular, I don't understand the purpose of the trigonometric substitutions as the integral should be solvable already without them. Maybe there was a reason in context. I assume the book had some reason for using that derivation.
Anyways, I've usually seen integration by parts written $$\big\[uv\big\]\_{x=t\_0}^{x=t} = \int\_{t\_0}^t u dv + \int\_{t\_0}^t v du$$ and usually solved for one of the integrals, e.g. $$\int\_{t\_0}^t v du = \big\[uv\big\]\_{x=t\_0}^{x=t} - \int\_{t\_0}^t u dv$$ where $\big[uv\big]\_{x=t\_0}^{x=t}$ means evaluate $uv$ at $t$ and subtract $uv$ evaluated at $t_0$, i.e. $\big[f(x)\big]\_{x=t\_0}^{x=t} = f(t) - f(t\_0)$. (This operation can actually be understood as a $0$-dimensional contour integral.) I'll abbreviate this as $\big[f(x)]\_{t\_0}^t$.
The derivation of this is quite simple by integrating the product rule, i.e. we have $$\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$ or often written $$d(uv) = udv + vdu$$ Integrating and applying the fundamental theorem of calculus gives you integration by parts as I've written it above.
For your specific example, I (personally) would choose $du = e^{ax}dx$ and $v = \cos(bx + c)$. (The other choice is equally valid, though in many cases one choice is significantly easier than the other. Here, it doesn't make that much of a difference.) I'll also set $t\_0 = 0$ as it only affects the result by a constant term. This gives $$\int\_0^t e^{ax}\cos(bx + c)dx = \left[\frac{1}{a}e^{ax}\cos(bx + c)\right]\_0^t + \frac{b}{a}\int\_0^t e^{ax}\sin(bx+c)dx$$
We now apply integration by parts again to the second integral and again choosing $du = e^{ax}dx$ and $v = \sin(bx + c)$ producing $$\int\_0^t e^{ax}\sin(bx+c)dx = \left[\frac{1}{a}e^{ax}\sin(bx + c)\right]\_0^t - \frac{b}{a}\int\_0^t e^{ax}\cos(bx + c)dx$$ We see have reproduced the integral we were originally trying to solve. This is a common pattern and usually a good thing. Sometimes the integral might be one of the intermediate steps. The important point is once we're in this situation, we now have an "algebraic" equation. This is indeed made clearer by naming the integral $I$, i.e. setting $I=\int\_0^t e^{ax}\cos(bx + c)dx$. Substituting this second integration by parts back into the first and using this substitution, we get $$I = \left[\frac{1}{a}e^{ax}\cos(bx + c)\right]\_0^t + \frac{b}{a}\left(\left[\frac{1}{a}e^{ax}\sin(bx + c)\right]\_0^t - \frac{b}{a}I\right)$$
The important point here is that, as an equation in $I$, this no longer involves any integration. Distributing the $\frac{b}{a}$ and multiplying the whole equation by $a^2$ gives $$a^2I = \left[ae^{ax}\cos(bx + c) ight]\_0^t + b\left[e^{ax}\sin(bx + c) ight]\_0^t - b^2I$$ We can easily solve for $I$ (the original integral) by adding $b^2I$ to both sides then dividing by $a^2 + b^2$. $$I = \frac{\left[ae^{ax}\cos(bx + c) ight]\_0^t + b\left[e^{ax}\sin(bx + c) ight]\_0^t}{a^2 + b^2}$$ Simplifying a bit produces $$I = \frac{ae^{at}\cos(bt + c) - a\cos(c) + be^{at}\sin(bt + c) - b\sin(c)}{a^2 + b^2}$$ The correctness of this expression can be verified by differentiating with respect to $t$ and evaluating at $t=0$ which should result in $0$.
(Connecting to your other question, we could rewrite $\cos(bx + c)$ as $(e^{i(bx + c)} + e^{-i(bx + c)})/2$ and the integral would split into two straightforward integrals of exponential functions.)

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