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Question:

The value of the integral ∫ln2ln3 [x sin(x²/2)] / [sin(x²/2) + sin(ln6 - x²/2)] dx is?

1/2 ln(3/2)

ln(3/2)

1/6 ln(3/2)

1/4 ln(3/2)

Solution:

Let I = ∫ln2ln3 [x sin(x²/2)] / [sin(x²/2) + sin(ln6 - x²/2)] dx
Now, substitute ln6 - x²/2 = t
Thus, when x = ln3 -> t = ln6 - ln3 = ln2
Also, when x = ln2 -> t = ln6 - ln2 = ln3
-> -2x dx = dt
Thus, -> dx = -dt / 2√(ln6 - t)
Substituting the above values in I, we get
I = ∫ln2ln3 √(ln6 - t) sin(ln6 - t) / [sin(t) + sin(ln6 - t)] (-dt / 2√(ln6 - t))
I = -1/2 ∫ln2ln3 sin(ln6 - t) / [sin(t) + sin(ln6 - t)] dt
I = 1/2 ∫ln3ln2 sin(ln6 - t) / [sin(t) + sin(ln6 - t)] dt.. (1)
Now also, ∫ab f(x) dx = ∫ab f(a + b - x) dx
Thus, expression (1) can also be written as,
I = 1/2 ∫ln3ln2 sin(t) / [sin(t) + sin(ln6 - t)] dt.. (2)
Adding equations (1) and (2), we get,
-> 2I = 1/2 ∫ln3ln2 [sin(ln6 - t) + sin(t)] / [sin(t) + sin(ln6 - t)] dt
-> 2I = 1/2 ∫ln3ln2 1 dt
-> 2I = 1/2 [t]ln3ln2
-> 2I = 1/2 [ln2 - ln3]
-> 2I = -1/2 ln(3/2)
-> I = -1/4 ln(3/2)
Therefore, I = 1/4 ln(3/2)