If I_n = int sin^n x , dx,then n I_n - (n-1) | vedclass

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(C) We are given the reduction formula for $I_n = \int \sin^n x \, dx$. Using integration by parts,let $u = \sin^{n-1} x$ and $dv = \sin x \, dx$. The

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$-\sin^{n-1} x \cos x$

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(C) We are given the reduction formula for $I_n = \int \sin^n x \, dx$. Using integration by parts,let $u = \sin^{n-1} x$ and $dv = \sin x \, dx$. Then $du = (n-1) \sin^{n-2} x \cos x \, dx$ and $v = -\cos x$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I_n = -\sin^{n-1} x \cos x - \int (-\cos x) (n-1) \sin^{n-2} x \cos x \, dx$ $I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \cos^2 x \, dx$ $I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x (1 - \sin^2 x) \, dx$ $I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2} - (n-1) I_n$ Rearranging the terms: $I_n + (n-1) I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$ $n I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$ Therefore,$n I_n - (n-1) I_{n-2} = -\sin^{n-1} x \cos x$.

If $I_n = \int \sin^n x \, dx$,then $n I_n - (n-1) I_{n-2}$ equals

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