If I_n = int_0^{frac{pi}{2}} cos^n x cos(nx) | vedclass

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(B) Given $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$. For $n=1$: $I_1 = \int_0^{\frac{\pi}{2}} \cos x \cos x dx = \int_0^{\frac{\pi}{2}} \fra

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$G$.$P$.

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(B) Given $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$. For $n=1$: $I_1 = \int_0^{\frac{\pi}{2}} \cos x \cos x dx = \int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2} dx = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}$. For $n=2$: $I_2 = \int_0^{\frac{\pi}{2}} \cos^2 x \cos 2x dx$. Using $\cos 2x = 2\cos^2 x - 1$, we have $\cos^2 x = \frac{1+\cos 2x}{2}$. $I_2 = \int_0^{\frac{\pi}{2}} \left( \frac{1+\cos 2x}{2} \right) \cos 2x dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos 2x dx + \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos^2 2x dx$. $I_2 = \frac{1}{2} \left[ \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{2}} + \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1+\cos 4x}{2} dx = 0 + \frac{1}{4} \left[ x + \frac{\sin 4x}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{8}$. For $n=3$: $I_3 = \int_0^{\frac{\pi}{2}} \cos^3 x \cos 3x dx$. Using $\cos 3x = 4\cos^3 x - 3\cos x$, we have $\cos^3 x = \frac{\cos 3x + 3\cos x}{4}$. $I_3 = \int_0^{\frac{\pi}{2}} \left( \frac{\cos 3x + 3\cos x}{4} \right) \cos 3x dx = \frac{1}{4} \int_0^{\frac{\pi}{2}} \cos^2 3x dx + \frac{3}{4} \int_0^{\frac{\pi}{2}} \cos x \cos 3x dx$. $I_3 = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{1+\cos 6x}{2} dx + \frac{3}{8} \int_0^{\frac{\pi}{2}} (\cos 4x + \cos 2x) dx = \frac{1}{8} [x]_0^{\frac{\pi}{2}} + 0 + 0 = \frac{\pi}{16}$. Since $I_1 = \frac{\pi}{4}$, $I_2 = \frac{\pi}{8}$, $I_3 = \frac{\pi}{16}$, the ratio $\frac{I_2}{I_1} = \frac{1}{2}$ and $\frac{I_3}{I_2} = \frac{1}{2}$. Thus, $I_1, I_2, I_3, \ldots$ are in $G$.$P$.

If $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$, then $I_1, I_2, I_3, \ldots$ are in

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