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AIEEE 2011

JEE Main 2011 Mathematics question on Definite Integrals and Areas

The value of ∫log⁡2log⁡3xsin⁡x2sin⁡x2+sin⁡(log⁡6−x2)dx\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2 + \sin(\log 6 - x^2)} dx is \hspace{1em}
  1. A.
    14log⁡32\frac{1}{4} \log \frac{3}{2}
  2. B.
    12log⁡32\frac{1}{2} \log \frac{3}{2}
  3. C.
    log⁡32\log \frac{3}{2}
  4. D.
    16log⁡32\frac{1}{6} \log \frac{3}{2}
Show answer and solution

Correct option: A

Solution

Substitute t = x², dt = 2x dx, so the integral becomes (1/2)∫[log2 to log3] sin(t)/(sin(t) + sin(log6 - t)) dt.

Apply the King property: for ∫[a to b] f(t)dt, replace t → (a+b-t). Here a+b = log2+log3 = log6, so t → log6-t, which swaps sin(t) and sin(log6-t) in numerator and denominator.

Calling the integral I, adding I with its transformed version gives 2I = (1/2)∫[log2 to log3] 1 dt = (1/2)(log3 - log2) = (1/2)log(3/2).

Therefore I = (1/4)log(3/2).

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