$$
\begin{aligned}
& \text { (a) } I=\int \frac{d x}{\cos ^4 x+\sin ^4 x} \\
& =\int \frac{\sec ^4 x}{1+\tan ^4 x} d x=\int \frac{\sec ^2 x\left(1+\tan ^2 x\right)}{1+\tan ^4 x} d x
\end{aligned}
$$

Let $\tan x=t$, then $\sec ^2 x d x=d t$
$$
\begin{aligned}
I & =\int \frac{1+t^2}{1+t^4} d t=\int \frac{\left(1+\frac{1}{t^2}\right)}{\left(t^2+\frac{1}{t^2}\right)} d t \\
& =\int \frac{\left(1+\left(1 / t^2\right)\right)}{\left(t-\frac{1}{t}\right)^2+2} d t
\end{aligned}
$$

Let $t-\frac{1}{t}=u$, then $\left(1+\frac{1}{t^2}\right) d t=d u$
$$
\begin{aligned}
I & =\int \frac{d u}{u^2+2}=\int \frac{d u}{u^2+(\sqrt{2})^2} \\
& =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+C \\
I & =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+C \\
& =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t^2-1}{\sqrt{2} t}\right)+C \\
I & =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan ^2 x-1}{\sqrt{2} \tan x}\right)+C \\
\therefore g(x) & =\frac{\tan ^2 x-1}{\sqrt{2} \tan x}=\frac{1}{\sqrt{2}}(\tan x-\cot x)
\end{aligned}
$$