Relation between $I_{k}$ and $I_{k+2}$ 
If $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,n\in \mathbb{N}$ Then Relation between $I_{k}$ and $I_{k+2}$

$\bf{My\; Try::}$ Given  $\displaystyle I_{k}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^ndx\;,$ Then $\displaystyle I_{k+2}=\int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^{n+2}dx\;,$
So $$I_{k+2}-I_{k} = \int_{0}^{\frac{\pi}{2}}x(\sin x+\cos x)^n\cdot \sin 2xdx$$
Now How can I solve it after that, Help me
Thanks


Note:
  $$\int_0^\frac{\pi}{2} x(\sin x+\cos x)^n dx=\frac{\pi}{4}\int_0^\frac{\pi}{2}(\sin x+\cos x)^ndx=I_n$$ 
  (Put $x=\frac{\pi}{2}-x$ to prove this)

We have $$I_{n+2}-I_n=\frac{\pi}{4}\int_0^\frac{\pi}{2} (\sin x+\cos x)^n \sin 2xdx$$
Applying by parts, differentiating $(\sin x+\cos x)^n$ and integrating $\sin 2x$, this becomes:
$$I_{n+2}-I_n=\frac{\pi}{4}\left (-\frac 12 \cos 2x(\sin x+\cos x)^n|_0^{\frac{\pi}{2}}+\frac{n}{2}\int_0^\frac{\pi}{2}(\sin x+\cos x)^n dx-\frac{n}{2}\int_0^\frac{\pi}{2}(\sin x+\cos x)^n \sin 2x dx\right)$$
Now use $$\int_0^\frac{\pi}{2}(\sin x+\cos x)^n dx=\frac{4I_n}{\pi}$$
and
$$\int_0^\frac{\pi}{2}(\sin x+\cos x)^n \sin 2x dx=\frac{4(I_{n+2}-I_n)}{\pi}$$
to get:
$$\color{green} {I_{n+2}=I_n+\frac{\pi}{2(n+2)}}$$
