$=\int_0^{\pi / 2} \frac{\sin ^3\left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x$

Adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I & =\int_0^{\pi / 2} \frac{\left(\sin ^3 x \cos x+\cos ^3 x \sin x\right)}{\cos ^4 x+\sin ^4 x} d x \\
& =\int_0^{\pi / 2} \frac{\sin x \cos x\left(\sin ^2 x+\cos ^2 x\right)}{\cos ^4 x+\sin ^4 x} d x \\
& =\int_0^{\pi / 2} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =\int_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^4 x\left(\tan ^4 x+1\right)} d x \\
& =\int_0^{\pi / 2} \frac{\tan ^4 \sec ^2 x}{\left(\tan ^4 x+1\right)} d x
\end{aligned}
$$

Let $\tan ^2 x=t$
Differentiating w.r.t. $x$, we get
$$
\begin{array}{ll}
& 2 \tan x \sec ^2 x d x=d t \\
\therefore & \tan x \sec ^2 x d x=\frac{d t}{2} \\
\Rightarrow & 2 I=\frac{1}{2} \int_0^{\infty} \frac{d t}{t^2+1}=\frac{1}{2}\left[\tan ^{-1} t\right]_0^{\infty} \\
\Rightarrow & 2 I=\frac{1}{2}\left[\tan ^{-1}(\infty)-\tan ^{-1}(0)\right]=\frac{1}{2} \times \frac{\pi}{2} \\
& 2 I=\frac{\pi}{4} \Rightarrow I=\frac{\pi}{8}
\end{array}
$$