$$
\begin{aligned}
& I_n=\int \cos ^n x d x \\
& =\int \cos ^{n t-1} x \cdot \cos x d x \\
& =\cos ^{n t-1} x \cdot \sin x-\int(n-1) \cos ^{n t-2} x \\
& (-\sin x) \cdot(\sin x) d x \\
& =\cos ^{n t-1} x \sin x+\int \cos ^{n t-2} x \cdot(n-1) \sin ^2 x d x \\
& =\cos ^{n+1} x \sin x+(n-1) \int \cos ^{n t-2} x\left(1-\cos ^2 x\right) d x \\
& =\cos ^{n-1} x \sin x+(n-1) \int \cos ^{n-2} x-(n-1) \\
& \int \cos ^n x d x \\
& =\cos ^{n-1} \chi \sin \chi+(n-1) I_{n-2}-(n-1) I_n \\
& \therefore I_n(1+n-1)=\cos ^{n-1} x \sin x+(n-1) Y_{n t}-2 \\
& \Rightarrow n I_{n t}-(n-1) I_{n-2}=\cos ^{n-1} x \sin x \\
&
\end{aligned}
$$
Putting $n=6$, we get
$$
6 I_6-5 I_4=\cos ^5 x \sin x
$$