Let the domain of the function f(x) = log_{4} | vedclass

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(C) For the function to be defined,the arguments of the logarithms must be positive:$\log_{5}(\log_{3}(18x - x^{2} - 77)) > 0 \implies \log_{3}(18x -

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$1$

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(C) For the function to be defined,the arguments of the logarithms must be positive:$\log_{5}(\log_{3}(18x - x^{2} - 77)) > 0 \implies \log_{3}(18x - x^{2} - 77) > 1 \implies 18x - x^{2} - 77 > 3$$x^{2} - 18x + 80 Thus,$a = 8$ and $b = 10$.Let $I = \int_{a}^{b} \frac{\sin^{3} x}{\sin^{3} x + \sin^{3}(a + b - x)} dx$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we get:$I = \int_{a}^{b} \frac{\sin^{3}(a + b - x)}{\sin^{3}(a + b - x) + \sin^{3} x} dx$.Adding the two expressions for $I$:$2I = \int_{a}^{b} \frac{\sin^{3} x + \sin^{3}(a + b - x)}{\sin^{3} x + \sin^{3}(a + b - x)} dx = \int_{a}^{b} 1 dx = b - a$.$I = \frac{b - a}{2} = \frac{10 - 8}{2} = 1$.

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 $.....$

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