Question

The value of \(\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin(\ln 6 - x^2)} dx\) is

• A. \(\frac{1}{4} \ln \frac{3}{2}\)
• B. \(\frac{1}{2} \ln \frac{3}{2}\)
• C. \(\ln \frac{3}{2}\)
• D. \(\frac{1}{6} \ln \frac{3}{2}\)

Hint:

Property: \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\) is very useful for integrals with symmetric limits.

Answer 1

The Correct Option is A

Let's evaluate the integral \( \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin(\ln 6 - x^2)} \, dx \). 

We can use the property of definite integrals for simplification. Consider the substitution \( x^2 = t \). Then \( 2x \, dx = dt \), or \( x \, dx = \frac{1}{2} \, dt \).

Changing the limits of integration accordingly, when \( x = \sqrt{\ln 2} \), \( t = \ln 2 \) and when \( x = \sqrt{\ln 3} \), \( t = \ln 3 \).

The integral becomes:

\(\int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin(\ln 6 - t)} \cdot \frac{1}{2} \, dt\)

This can be rewritten using the given symmetrical property:

\(I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin(\ln 6 - t)} \, dt\)

Using symmetry, it's known that:

\(\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx\)

For our integral:

\(J = \int_{\ln 2}^{\ln 3} \frac{\sin(\ln 6 - t)}{\sin(\ln 6 - t) + \sin t} \, dt\)

Notice that \(I = J\) due to the symmetrical property, because changing variables from \( t \to \ln 6 - t \) gives the same integral:

\(I + J = \int_{\ln 2}^{\ln 3} 1 \, dt = \ln 3 - \ln 2 = \ln \frac{3}{2}\)

Therefore, \(2I = \ln \frac{3}{2}\), which results in:

\(I = \frac{1}{2} \ln \frac{3}{2}\)

The value of the original integral is:

\(\frac{1}{4} \ln \frac{3}{2}\)

Thus, the correct answer is \(\frac{1}{4} \ln \frac{3}{2}\).