JEE Main 2025 — 7 April, Morning Shift — Question 31

$I=\int_{0}^{\pi} \frac{(x+3) \sin x}{1+3 \cos ^{2} x} d x …(i)$

$=\int_{0}^{\pi} \frac{(\pi-x+3) \sin (\pi-x)}{1+3 \cos ^{2}(\pi-x)} d x$

$$\begin{gathered} =\int_{0}^{\pi} \frac{(\pi+3) \sin x-x \sin x}{1+3 \cos ^{2} x} d x …(ii) \end{gathered}$$

Add (i) & (ii) $2 I=\int_{0}^{\pi} \frac{(\pi+6) \sin x}{1+3 \cos ^{2} x} d x$

Let $\cos x=t \Rightarrow-\sin x d x=d t$

$(\pi+6) \int_{1}^{-1} \frac{-d t}{\left(1+3 t^{2}\right)}=\left(\frac{\pi+6}{3}\right) \int_{-1}^{1} \frac{d t}{t^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}}$

$\Rightarrow \quad 2 I=\left.\left(\frac{\pi+6}{3}\right) \cdot \frac{1}{\left(\frac{1}{\sqrt{3}}\right)} \cdot \tan ^{-1} \frac{t}{\frac{1}{\sqrt{3}}}\right|_{-1} ^{1}$

$$\begin{aligned} & \Rightarrow \quad 2 I=\left(\frac{\pi+6}{\sqrt{3}}\right)\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(-\sqrt{3})\right] \\&=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right) \\&=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot \frac{2 \pi}{3} \\& \Rightarrow \quad I=\frac{\pi(\pi+6)}{3 \sqrt{3}} \end{aligned}$$