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Step-by-Step Solution
Step 1: Find the Domain of the Given Function
We are given
f(x) = \log_{4}\bigl(\log_{5}(\log_{3}(18x - x^2 - 77))\bigr).
For f(x) to be defined, every logarithm inside must be positive:
The innermost expression 18x - x^2 - 77 must be greater than 1 (because of the subsequent steps of the logarithm, explained below).
First, we need \log_{3}(18x - x^2 - 77) > 0 so that it can be valid for the next log base 5. But actually, as given in the solution, we need a bit stronger condition:
\log_{5}\bigl(\log_{3}(18x - x^2 - 77)\bigr) > 0.
This implies
\log_{3}(18x - x^2 - 77) > 1.
Therefore,
18x - x^2 - 77 > 3 \quad \Longrightarrow \quad x^2 - 18x + 80 < 0.
Solving x^2 - 18x + 80 < 0 gives the interval (8, 10).
Hence, the domain of f(x) is (8, 10) . Thus, we identify a = 8 and b = 10 .
Step 2: Set Up the Integral
We need to evaluate the definite integral
I = \int_{a}^{b} \frac{\sin^3 x}{\sin^3 x + \sin^3(a + b - x)} \, dx,
where a = 8 and b = 10 .
Step 3: Use the Integral Property by Symmetry
A common technique for integrals of this form is to perform the substitution x \mapsto (a + b - x) and observe what happens to the integrand.
Under x \mapsto (a + b - x) , the integral becomes
\int_{a}^{b} \frac{\sin^3(a + b - x)}{\sin^3(a + b - x) + \sin^3 x} \, dx.
Notice that
\frac{\sin^3 x}{\sin^3 x + \sin^3(a + b - x)}
\;+\;
\frac{\sin^3(a + b - x)}{\sin^3 x + \sin^3(a + b - x)}
= 1.
Hence, if we call the original integral I , then after substituting and adding both integrals, we effectively get
I + I = \int_{a}^{b} 1 \, dx = (b - a).
Step 4: Solve for the Integral
From the above reasoning,
2I = (b - a).
Therefore,
I = \frac{b - a}{2}.
Since a = 8 and b = 10 , we have
I = \frac{10 - 8}{2} = \frac{2}{2} = 1.
Step 5: Final Answer
The value of the integral
\int_{8}^{10} \frac{\sin^3 x}{\sin^3 x + \sin^3(8 + 10 - x)} \, dx
is
\boxed{1}.
