The integral $\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x$ is equal to

<p>$$\begin{aligned} & I=\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (i)}\\ & =\int_0^\pi \frac{(\pi-x+3) \sin (\pi-x)}{1+3 \cos ^2(\pi-x)} d x \\ & =\int_0^\pi \frac{(\pi+3) \sin x-x \sin x}{1+3 \cos ^2 x} d x \quad\text{..... (ii)} \end{aligned}$$</p> <p>Add (i) & (ii)</p> <p>$2 I=\int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos ^2 x} d x$</p> <p>Let $\cos x=t \Rightarrow-\sin x d x=d t$</p> <p>$(\pi+6) \int_1^{-1} \frac{-d t}{\left(1+3 t^2\right)}=\left(\frac{\pi+6}{3}\right) \int_{-1}^1 \frac{d t}{t^2+\left(\frac{1}{\sqrt{3}}\right)^2}$</p> <p>$\Rightarrow \quad 2 I=\left.\left(\frac{\pi+6}{3}\right) \cdot \frac{1}{\left(\frac{1}{\sqrt{3}}\right)} \cdot \tan ^{-1} \frac{t}{\frac{1}{\sqrt{3}}}\right|_{-1} ^1$</p> <p>$$\begin{aligned} & \Rightarrow \quad 2 I=\left(\frac{\pi+6}{\sqrt{3}}\right)\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(-\sqrt{3})\right] \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right) \\ &=\left(\frac{\pi+6}{\sqrt{3}}\right) \cdot \frac{2 \pi}{3} \\ & \Rightarrow \quad I=\frac{\pi(\pi+6)}{3 \sqrt{3}} \end{aligned}$$</p>