For n ≥ 2 , if I n = ∫ ( sin x + cos x ) n d x , then n I n - 2 ( n - 1 ) I n - 2 =

The integral expression is given as,

$I_n=\int (\sin x+\cos x{)}^{n}dx$

$\Rightarrow I_n=\int (\sin x+\cos x{)}^{n-1}·(\sin x+\cos x)dx$

As we know from integration by parts $\int uvdx=u\int vdx-\int \left\{\frac{du}{dx}·\int vdx\right\}dx$

$\Rightarrow I_n=(\sin x+\cos x{)}^{n-1} (\sin x-\cos x)-(n-1)\int (\sin x+\cos x{)}^{n-2}\left(\cos x-\sin x\right)·(\sin x-\cos x)dx$

$\left\{\int \sin xdx=-\cos x, \int \cos xdx=\sin x\right\}$

$\Rightarrow I_n=(\sin x+\cos x{)}^{n-1}(\sin x-\cos x)-(n-1)\int (\sin x+\cos x{)}^{n-2}\left(-1+\sin 2x\right)dx+c_1$

$\Rightarrow I_n=(\sin x+\cos x{)}^{n-1}(\sin x-\cos x)+(n-1)\int (\sin x+\cos x{)}^{n-2} dx-(n-1)\int (\sin x+\cos x{)}^{n-2}\left(\sin 2x\right)dx+c_1$

$\Rightarrow I_n=(\sin x+\cos x{)}^{n-1}(\sin x-\cos x)+2(n-1)\int (\sin x+\cos x{)}^{n-2}dx-(n-1)\int (\sin x+\cos x{)}^{n-2}\left(1+\sin 2x\right)dx+c_1$

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

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