Let the domain of the function $f(x) = \log_4(\log_5(\log_3(18x-x^2-77)))$ be $(a, b)$. Then the value of the integral $\int_a^b \frac{\sin^3 x}{\sin^3 x + \sin^3(a+b-x)} dx$ is equal to _________.
Show Hint
Recognize the standard integral form $\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} dx$. By applying the King's property ($\int_a^b g(x)dx = \int_a^b g(a+b-x)dx$) and adding the original and transformed integrals, the result is always $\frac{b-a}{2}$. In this case, $\frac{10-8}{2} = 1$.
Step 1: Find the domain of the function
For the given function to be defined, the argument of each logarithm must be positive.
\[
\log_4(\cdot) \text{ defined } \Rightarrow \log_5(\log_3(18x-x^2-77))>0
\]
\[
\Rightarrow \log_3(18x-x^2-77)>1
\]
\[
\Rightarrow 18x-x^2-77>3
\]
\[
\Rightarrow 18x-x^2-80>0
\]
\[
\Rightarrow x^2-18x+80<0
\]
Factoring:
\[
(x-8)(x-10)<0
\]
Hence,
\[
8<x<10
\]
So,
\[
(a,b)=(8,10)
\]
Step 2: Evaluate the integral
Let
\[
I=\int_8^{10}\frac{\sin^3 x}{\sin^3 x+\sin^3(18-x)}\,dx
\]
Using the property of definite integrals:
\[
\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx
\]
\[
I=\int_8^{10}\frac{\sin^3(18-x)}{\sin^3(18-x)+\sin^3 x}\,dx
\]
Step 3: Add both expressions
\[
2I=\int_8^{10}\left(
\frac{\sin^3 x}{\sin^3 x+\sin^3(18-x)}
+
\frac{\sin^3(18-x)}{\sin^3(18-x)+\sin^3 x}
\right)dx
\]
\[
2I=\int_8^{10}1\,dx
\]
\[
2I=10-8=2
\]
\[
I=1
\]
Answer: \(\boxed{1}\)
