Answer
The second integral requires the reduction formula and more work to evaluate.
Work Step by Step
Given $$\int \sin ^{798} x \cos x d x$$
Here, we use simple substitution:
$$u= \sin x\ \ \ \ \ \ du=\cos xdx $$
Then
\begin{align*}
\int \sin ^{798} x \cos x d x&=\int u^{798} d u\\
&= \frac{1}{799}u^{799}+C\\
&= \frac{1}{799}(\sin x)^{799}+C\\
\end{align*}
For $$\int \sin ^{4} x \cos ^{4} x d x $$
Use the reduction formula:
$$ \int \sin ^{m} x \cos ^{n} x d x=\frac{\sin ^{m+1} x \cos ^{n-1} x}{m+n}+\frac{n-1}{m+n} \int \sin ^{m} x \cos ^{n-2} x d x$$
and
$$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x $$
Then
\begin{align*}
\int \sin ^{4} x \cos ^{4} x d x &= \frac{\sin ^{5} x \cos ^{3} x}{8}+\frac{3}{6} \int \sin ^{4} x \cos ^{2} x d x\\
&=\frac{\sin ^{5} x \cos ^{3} x}{8}+\frac{3}{6}\left(\frac{\sin ^{4 }x \cos x}{6}+\frac{1}{6} \int \sin ^{4} x d x\right) \\
&= \frac{\sin ^{5} x \cos ^{3} x}{8}+\frac{3}{6}\left(\frac{\sin ^{4 }x \cos x}{6}+\frac{1}{6} \left[-\frac{1}{4}\sin ^3\left(x\right)\cos \left(x\right)+\frac{3}{8}\left(x-\frac{1}{2}\sin \left(2x\right)\right)\right]\right)
\end{align*}
