For n ge 2,if I_n = int (sin x + cos x)^n dx, | vedclass

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(C) Let $I_n = \int (\sin x + \cos x)^n dx$. Using integration by parts,let $u = (\sin x + \cos x)^{n-1}$ and $dv = (\sin x + \cos x) dx$. Then $du =

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$(\sin x + \cos x)^{n-1}(\sin x - \cos x) + C$

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(C) Let $I_n = \int (\sin x + \cos x)^n dx$. Using integration by parts,let $u = (\sin x + \cos x)^{n-1}$ and $dv = (\sin x + \cos x) dx$. Then $du = (n-1)(\sin x + \cos x)^{n-2}(\cos x - \sin x) dx$ and $v = (-\cos x + \sin x) = (\sin x - \cos x)$. $I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - \int (n-1)(\sin x + \cos x)^{n-2}(\cos x - \sin x)(\sin x - \cos x) dx$. Since $(\cos x - \sin x)(\sin x - \cos x) = -(\sin x - \cos x)^2 = -(1 - \sin 2x) = \sin 2x - 1$. Also,$(\sin x + \cos x)^2 = 1 + \sin 2x$,so $\sin 2x = (\sin x + \cos x)^2 - 1$. Thus,$\sin 2x - 1 = (\sin x + \cos x)^2 - 2$. $I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - (n-1) \int (\sin x + \cos x)^{n-2} ((\sin x + \cos x)^2 - 2) dx$. $I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) - (n-1) I_n + 2(n-1) I_{n-2}$. $I_n + (n-1) I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + 2(n-1) I_{n-2}$. $n I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + 2(n-1) I_{n-2}$. $n I_n - 2(n-1) I_{n-2} = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + C$.

For $n \ge 2$,if $I_n = \int (\sin x + \cos x)^n dx$,then $nI_n - 2(n-1)I_{n-2} = $

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