JEE Mathematics 2025 Question with Solution

Given:

$$I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 \frac{3}{2}x}{\sin^2 x + \cos^2 x} \, dx$$

and we need to evaluate

$$J = \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$$

Find: The value of $J$.

From the solution, the key observations used are:

$$\sin^2 x + \cos^2 x = 1$$

so the first integral becomes

$$I = \int_0^{\frac{\pi}{2}} \sin^2 \frac{3x}{2} \, dx$$

The provided solution states that this evaluates to

$$I = \frac{\pi^2}{16}$$

For the required integral, use

$$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$$

and with

$$\sin 2x = 2\sin x \cos x$$

we get

$$\sin^2 x \cos^2 x = \frac{1}{4}\sin^2 2x$$

Hence,

$$\sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x$$

The provided solution then invokes symmetry by taking

$$u = \frac{\pi}{2} - x$$

and concludes that the required integral has value

$$J = \frac{\pi^2}{16}$$

Therefore, the correct option is A.

Given:

$$J = \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx$$

Find: The value of the integral.

The extracted solution rewrites the denominator as

$$\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$$

Then using

$$\sin^2 x \cos^2 x = \frac{1}{4}\sin^2 2x$$

it becomes

$$\sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x$$

So the integral is rewritten in a simplified trigonometric form. The solution then uses the reflection substitution

$$u = \frac{\pi}{2} - x$$

to exploit symmetry on the interval $\left[0, \frac{\pi}{2}\right]$. Based on this symmetry argument and the stated relation with the first integral, the final result reported on the page is

$$\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx = \frac{\pi^2}{16}$$

Thus, the correct option is A.

The solution is concise and concludes directly with the final value, so the answer is taken from that conclusion.