Definite Integration – Properties – JEE Main 2025 Shift 2

Solution:

First, calculate the value of $I$:
$$I = \int_{0}^{\pi/2} \frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x} dx$$
Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$:
$$I = \int_{0}^{\pi/2} \frac{\cos^{3/2}x}{\cos^{3/2}x+\sin^{3/2}x} dx$$
Adding both equations:
$$2I = \int_{0}^{\pi/2} 1 dx = [x]_0^{\pi/2} = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4}$$

Now evaluate the required integral with upper limit $2I = 2(\frac{\pi}{4}) = \frac{\pi}{2}$.
Let $J = \int_{0}^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx$.
Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$:
$$J = \int_{0}^{\pi/2} \frac{(\frac{\pi}{2}-x) \cos x \sin x}{\cos^4 x + \sin^4 x} dx$$

Adding the two expressions for $J$:
$$2J = \int_{0}^{\pi/2} \frac{\frac{\pi}{2} \sin x \cos x}{\sin^4 x + \cos^4 x} dx$$
$$2J = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$$
Divide numerator and denominator by $\cos^4 x$:
$$2J = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx$$

Let $\tan^2 x = t \Rightarrow 2 \tan x \sec^2 x dx = dt \Rightarrow \tan x \sec^2 x dx = \frac{dt}{2}$.
Limits: $x \to 0, t \to 0$; $x \to \frac{\pi}{2}, t \to \infty$.
$$2J = \frac{\pi}{2} \int_{0}^{\infty} \frac{dt/2}{t^2 + 1} = \frac{\pi}{4} [\tan^{-1} t]_0^{\infty}$$
$$2J = \frac{\pi}{4} \left( \frac{\pi}{2} – 0 \right) = \frac{\pi^2}{8}$$
$$J = \frac{\pi^2}{16}$$

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