If I_n = int cos^n x , dx, then 6 I_6 - 5 I_4 | vedclass

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(D) We use the reduction formula for $I_n = \int \cos^n x \, dx$. Integrating by parts, we have: $I_n = \int \cos^{n-1} x \cdot \cos x \, dx$ $= \cos^

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$\cos^5 x \sin x$

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(D) We use the reduction formula for $I_n = \int \cos^n x \, dx$. Integrating by parts, we have: $I_n = \int \cos^{n-1} x \cdot \cos x \, dx$ $= \cos^{n-1} x \sin x - \int (n-1) \cos^{n-2} x (-\sin x) \sin x \, dx$ $= \cos^{n-1} x \sin x + (n-1) \int \cos^{n-2} x (1 - \cos^2 x) \, dx$ $= \cos^{n-1} x \sin x + (n-1) I_{n-2} - (n-1) I_n$ Rearranging the terms: $I_n + (n-1) I_n = \cos^{n-1} x \sin x + (n-1) I_{n-2}$ $n I_n = \cos^{n-1} x \sin x + (n-1) I_{n-2}$ $n I_n - (n-1) I_{n-2} = \cos^{n-1} x \sin x$ For $n = 6$, we substitute into the formula: $6 I_6 - 5 I_4 = \cos^{6-1} x \sin x$ $6 I_6 - 5 I_4 = \cos^5 x \sin x$

If $I_n = \int \cos^n x \, dx$, then $6 I_6 - 5 I_4 = $

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