Recurrence relation for $I_n=\int (\sin x+\cos x)^n \,\Bbb dx$

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 $$n I_n = (\sin x + \cos x)^{n-1}(\sin x - \cos x) +2(n-1)I_{n-2}$$

Answer 2

Multiplying both sides of the identity

$$(\sin x +\cos x)^2=-(\sin x-\cos x)^2+2$$

with $n(\sin x+\cos x)^{n-2}$ we have

$$n(\sin x +\cos x)^n= -n(\sin x+\cos x)^{n-2}(\sin x-\cos x)^2+2n(\sin x+\cos x)^{n-2}.$$

Arranging the right hand side we have

$$(nI_n)'=\color{blue}{(-n+1)(\sin x+\cos x)^{n-2}(\sin x-\cos x)^2}\color{red}{-(\sin x+\cos x)^{n-2}(\sin x-\cos x)^2 +2(\sin x+\cos x)^{n-2}}+2(n-1)(\sin x+\cos x)^{n-2}$$

and

$$(nI_n)'=\color{blue}{(n-1)(\sin x+\cos x)^{n-2}(\cos x-\sin x)(\sin x-\cos x)}+\color{red}{(\sin x+\cos x)^{n-1}(\cos x +\sin x)}+(2(n-1)I_{n-2})'$$

and

$$(nI_n)'=(\color{purple}{(\sin x+\cos x)^{n-1}(\sin x-\cos x)})'+(2(n-1)I_{n-2})'.$$

Finally integrating we obtain the recurrence relation

$$nI_n=(\sin x+\cos x)^{n-1}(\sin x-\cos x)+2(n-1)I_{n-2}.$$