JEE Mathematics 2025 Question with Solution

Given:

$$I = \int_0^\pi \frac{(x + 3) \sin x}{1 + 3\cos^2 x} \, dx$$

Find: The value of the integral and the correct option.

Using King’s Rule,

$$\int_0^\pi f(x)\,dx = \int_0^\pi f(\pi-x)\,dx$$

for the integral

$$I = \int_0^\pi \frac{(x + 3) \sin x}{1 + 3\cos^2 x} \, dx.$$

Replacing $x$ by $\pi-x$,

$$I = \int_0^\pi \frac{(\pi - x + 3)\sin x}{1 + 3\cos^2 x} \, dx$$

Adding the two expressions,

$$2I = \int_0^\pi \frac{(x+3 + \pi - x + 3)\sin x}{1 + 3\cos^2 x} \, dx$$ $$2I = (\pi + 6)\int_0^\pi \frac{\sin x}{1 + 3\cos^2 x} \, dx$$

So,

$$I = \frac{\pi + 6}{2}\int_0^\pi \frac{\sin x}{1 + 3\cos^2 x} \, dx$$

Now let

$$\cos x = t$$

then

$$-\sin x \, dx = dt$$

When $x=0$, $t=1$ and when $x=\pi$, $t=-1$. Hence,

$$I = \frac{\pi + 6}{2}\int_1^{-1} \frac{-dt}{1 + 3t^2}$$ $$I = \frac{\pi + 6}{2}\int_{-1}^{1} \frac{dt}{1 + 3t^2}$$

Since the integrand is even,

$$I = (\pi + 6)\int_0^1 \frac{dt}{1 + 3t^2}$$

Now,

$$\int \frac{dt}{1+3t^2} = \frac{1}{\sqrt{3}}\tan^{-1}(\sqrt{3}t)$$

Therefore,

$$I = (\pi + 6)\left[\frac{\tan^{-1}(\sqrt{3}t)}{\sqrt{3}}\right]_0^1$$ $$I = \frac{\pi + 6}{\sqrt{3}}\left(\frac{\pi}{3}\right)$$ $$I = \frac{\pi(\pi + 6)}{3\sqrt{3}}$$

Therefore, the value of the integral is $\frac{\pi}{3\sqrt{3}}(\pi + 6)$, so the correct option is C.

Given:

$$I = \int_0^\pi \frac{(x + 3) \sin x}{1 + 3\cos^2 x} \, dx$$

Find: A consistent evaluation from the solution.

The first approach on the page splits the integral into two parts, but the listed intermediate values for those parts are inconsistent with the final answer. The reliable working shown in the second approach gives:

$$I = \frac{\pi + 6}{2}\int_0^\pi \frac{\sin x}{1 + 3\cos^2 x} \, dx$$

and then evaluates the remaining integral correctly by substitution.

This yields

$$I = \frac{\pi(\pi + 6)}{3\sqrt{3}}$$

which matches option C.

Therefore, the first approach contains a discrepancy in its intermediate statements, while the second approach correctly establishes the answer.