Hint : \(I_n=\int_0^{\pi / 2} \cos ^n x \cdot \cos n x d x\)
\(I_1=\int_0^{\pi / 2} \cos x \cdot \cos x d x=\int_0^{\pi / 2}\left(\frac{1+\cos 2 x}{2}\right) d x=\frac{\pi}{4}\)
\(I_2=\int_0^{\pi / 2} \cos ^2 x \cdot \cos 2 x d x=\int_0^{\pi / 2}-\sin ^2 x \cdot \cos 2 x d x\)
\(2 I_2=\int_0^{\pi / 2} \cos ^2 2 x=\int_0^{\pi / 2} \frac{1+\cos 4 x}{2} d x=\frac{\pi}{4}\)
\(\therefore \mathrm{I}_2=\frac{\pi}{8}\)
\(I_3=\int_0^{\pi / 2} \cos ^3 x \cdot \cos 3 x d x=\int_0^{\pi / 2}\left(\frac{3 \cos x+\cos 3 x}{4}\right) \cos 3 x d x\)
\(\begin{aligned}
& =\frac{3}{4} \int_0^{\pi / 2} \cos x \cdot \cos 3 x d x+\frac{1}{4} \int_0^{\pi / 2} \cos ^2 3 x d x \\
& =\frac{3}{8} \int_0^{\pi / 2}(\cos 2 x+\cos 4 x) d x+\frac{1}{4} \int_0^{\pi / 2} \frac{1+\cos 6 x}{2} d x \\
& =\frac{\pi}{16}
\end{aligned}\)
\(\therefore \mathrm{I}_1, \mathrm{I}_2, \mathrm{I}_3\) are in G.P.