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

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(B) આપેલ છે $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$. $n=1$ માટે: $I_1 = \int_0^{\frac{\pi}{2}} \cos x \cos x dx = \int_0^{\frac{\pi}{2}} \

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$G$.$P$.

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(B) આપેલ છે $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$. $n=1$ માટે: $I_1 = \int_0^{\frac{\pi}{2}} \cos x \cos x dx = \int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2} dx = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}$. $n=2$ માટે: $I_2 = \int_0^{\frac{\pi}{2}} \cos^2 x \cos 2x dx$. $\cos 2x = 2\cos^2 x - 1$ નો ઉપયોગ કરતા,$\cos^2 x = \frac{1+\cos 2x}{2}$. $I_2 = \int_0^{\frac{\pi}{2}} \left( \frac{1+\cos 2x}{2} \right) \cos 2x dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos 2x dx + \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos^2 2x dx$. $I_2 = \frac{1}{2} \left[ \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{2}} + \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1+\cos 4x}{2} dx = 0 + \frac{1}{4} \left[ x + \frac{\sin 4x}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{8}$. $n=3$ માટે: $I_3 = \int_0^{\frac{\pi}{2}} \cos^3 x \cos 3x dx$. $\cos 3x = 4\cos^3 x - 3\cos x$ નો ઉપયોગ કરતા,$\cos^3 x = \frac{\cos 3x + 3\cos x}{4}$. $I_3 = \int_0^{\frac{\pi}{2}} \left( \frac{\cos 3x + 3\cos x}{4} \right) \cos 3x dx = \frac{1}{4} \int_0^{\frac{\pi}{2}} \cos^2 3x dx + \frac{3}{4} \int_0^{\frac{\pi}{2}} \cos x \cos 3x dx$. $I_3 = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{1+\cos 6x}{2} dx + \frac{3}{8} \int_0^{\frac{\pi}{2}} (\cos 4x + \cos 2x) dx = \frac{1}{8} [x]_0^{\frac{\pi}{2}} + 0 + 0 = \frac{\pi}{16}$. અહીં $I_1 = \frac{\pi}{4}$,$I_2 = \frac{\pi}{8}$,$I_3 = \frac{\pi}{16}$ હોવાથી,ગુણોત્તર $\frac{I_2}{I_1} = \frac{1}{2}$ અને $\frac{I_3}{I_2} = \frac{1}{2}$ છે. તેથી,$I_1, I_2, I_3, \ldots$ એ $G$.$P$. માં છે.

જો $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$ હોય,તો $I_1, I_2, I_3, \ldots$ શેમાં છે?

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