Summation and integration $$\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin x|}{1+\pi^x}\,dx=a((n+b)!-c!),$$where $a,b,c\in \Bbb N$. Find the value of $a+b+c$.
My attempt
Let $$I=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin x|}{1+\pi^x}\,dx.\tag{1}$$
Apply $\int\limits_{a}^{b}f(x)\,dx=\int\limits_{a}^{b}f(a+b-x)\,dx$ to get
$$I=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\frac{|\sin(-x)|}{1+\pi^{-x}}\,dx.\tag{2}$$
Add $(1)$ and $(2)$,
$$2I=\sum_{r=1}^n\int_{-r(r!)}^{r(r!)}\left| \sin x\right|\,dx=2\sum_{r=1}^n\int_{0}^{r(r!)}| \sin x|\,dx$$
$$I=\sum_{r=1}^n\int_{0}^{r(r!)}\sin x \,dx$$
I could not solve further and factorials in the answer are confusing me.
 A: The problem as it stands is not consistent. If we take the case $n=1$ we get
$$\int_{-1}^1\frac{|\sin(x)|}{1+\pi^x}\,{\rm d}x = a((b+1)!-c!)$$
The right hand side is supposed to be an integer while the left hand side evaluates to $\simeq 0.4596$.
We can make the problem consistent by making a very simple change, namely the replacement $$r(r!) \to \pi r(r!)$$ This makes it plausible that there is a typo and that this is the 'true' problem that the problem-creator meant to ask. However, there might be other ways to do it so I can't say for sure. For example changing $\sin(x)\to \pi\sin(\pi x)$ which was mentioned by Jack in the comments above also works and is in fact equivalent to the $\pi r(r!)$ replacement.

With the change $r(r!) \to \pi r (r!)$ to the problem statement it becomes easy to solve since you have already done the tricky part of this problem which is to use a clever substitution to reduce the sum down to $\sum_{r=1}^n I(r)$ where 
$$I(r) = \int_0^{\pi r(r!)}|\sin(x)|\,{\rm d} x$$
This integral evaluates to $I(r) = r(r!)\int_0^{\pi}|\sin(x)|\,{\rm d} x = 2r(r!)$ and it follows that
$$\sum_{r=1}^n I(r) = \sum_{r=1}^n 2[(r+1)! - r!] = 2[(n+1)!-1] \implies a+b+c = 4$$
