Let $$I=\int \frac{\sin x}{3+4 \cos ^2 x} d x$$

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

$\therefore\quad I=-\int \frac{d t}{3+4 t^2}=-\frac{1}{4} \int \frac{d t}{\left(\frac{\sqrt{3}}{2}\right)^2+t^2}$

$$\begin{aligned} & =-\frac{1}{4} \cdot \frac{2}{\sqrt{3}} \tan ^{-1} \frac{2 t}{\sqrt{3}}+C \\ & =-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C \end{aligned}$$

We have, $$\begin{aligned} f(x) & =\int_{-\pi}^\pi \sin ^3 x \cos ^2 x d x \\ f(-x) & =\int_{-\pi}^\pi \sin ^3(-2)-\cos ^2(-x) d x \\ & =-f(x) \end{aligned}$$

$$\begin{aligned} &\text { Since, } f(x) \text { is an odd function. }\\ &\therefore \quad \int_{-\pi}^\pi \sin ^3 x \cos ^2 x d x=0 \end{aligned}$$