Integration – Recurrence Relation for Integral of (sin x + cos x)^n

Question

We can obtain a recurrence relation for
$$I_n=\int (\sin x+\cos x)^n \,\Bbb dx$$
as $$nI_n= (\sin x+ \cos x)^{n-1} (\sin x- \cos x)+2(n-1)I_{n-2} \tag{$\ast$}$$
By writing $I_n=2^{n/2}J_n$, where $J_n=\int \sin^n (x+a) \,\Bbb dx=\int \sin ^n t \,\Bbb dt$
\begin{align} J_n&=\int \sin ^{n-2} t (1-\cos^2t) \,\Bbb dt \\ &=J_{n-2}-\int \cos t ~ \sin^{n-2}t ~\cos t \,\Bbb dt \end{align}
Integration by parts yields
$$J_n=J_{n-2}-\cos t~ \frac{\sin^{n-1} t}{n-1}-\frac{1}{n-1}J_n\implies nJ_n=(n-1)J_n-\cos t ~\sin^{n-1} t.$$
The question is: How else can one prove/get $(\ast)$?

Answer 1

You can write $$I_n = \int (\sin x + \cos x)^n \,dx = \int (\sin x + \cos x)^{n-1}(\sin x - \cos x)'\,dx$$ and integrate by parts to get $$I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) + (n-1)\int(1-2\sin x \cos x)(\sin x + \cos x)^{n-2}\,dx$$ Now write $$1-2\sin x \cos x = 2 - (\sin x + \cos x)^2$$ and substitute that into the last expression to get $$n I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) +2(n-1)I_{n-2}$$