Let $I=\int_0^a \frac{1}{1+4 x^2} d x=\frac{\pi}{8}$

Now, $\quad \int_0^a \frac{1}{4\left(\frac{1}{4}+x^2\right)} d x=\frac{2}{4}\left[\tan ^{-1} 2 x\right]_0^a$

$=\frac{1}{2} \tan ^{-1} 2 a-0=\pi / 8$

$$\begin{aligned} \frac{1}{2} \tan ^{-1} 2 a & =\frac{\pi}{8} \\ \Rightarrow\quad\tan ^{-1} 2 a & =\pi / 4 \\ \Rightarrow\quad 2 a & =1 \\ \therefore\quad a & =\frac{1}{2} \end{aligned}$$

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

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