Let the domain of the function $f(x)=\log _{4}\left(\log _{5}\left(\log _{3}\left(18 x-x^{2}-77\right)\right)\right)$ be $(a, b)$. Then the value of the integral $\int_{a}^{b} \frac{\sin ^{3} x}{\left(\sin ^{3} x+\sin ^{3}(a+b-x)\right)} d x$ is equal to ____.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Definite Integrals

Let the domain of the function

f (x) = lo g_{4} (lo g_{5} (lo g_{3} (18 x - x^{2} - 77)))

(a, b)

.
Then the value of the integral

\int_{a}^{b} \frac{s i n ^{3} x}{( s i n ^{3} x + s i n ^{3} ( a + b - x ) )} d x

is equal to ____.

Answer: 1

The outer logarithm requires

lo g_{5} (lo g_{3} (18 x - x^{2} - 77)) > 0

, so

lo g_{3} (18 x - x^{2} - 77) > 1

, which gives

18 x - x^{2} - 77 > 3 \Rightarrow x^{2} - 18 x + 80 < 0 \Rightarrow (x - 8) (x - 10) < 0

x \in (8, 10)

. On this interval

18 x - x^{2} - 77 > 3 > 0

, so both inner logarithms are defined and the domain is

(a, b) = (8, 10)

, with

a + b = 18

.
Writing

I = \int_{8}^{10} \frac{sin ^{3} x}{sin ^{3} x + sin ^{3} ( 18 - x )} d x

and replacing

x

18 - x

I = \int_{8}^{10} \frac{sin ^{3} ( 18 - x )}{sin ^{3} ( 18 - x ) + sin ^{3} x} d x

Adding the two forms,

2 I = \int_{8}^{10} d x = 2

, so the value of the integral is

1

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer