A Definite Integral

$\begin{aligned} & I_n = \int_0^{\pi/2}x(\sin(x)+\cos(x))^n\,dx \\ & I_n =\frac{1}{2} \int_0^{\pi/2}\left ( x(\sin(x)+\cos(x))^n + (\frac{\pi}{2}-x)(\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x))^n \right )\,dx && {\color{#D61F06} \int_a^b f(x)dx=\int_a^b f(a+b-x)dx}\\ & I_n =\frac{1}{2} \int_0^{\pi/2}\left ( x(\sin(x)+\cos(x))^n + (\frac{\pi}{2}-x)(\cos(x)+\sin(x))^n \right )\,dx && {\color{#D61F06} \sin(\frac{\pi}{2}-x)=\cos(x),\cos(\frac{\pi}{2}-x)=\sin(x)}\\ & I_n = \frac{1}{2} \int_0^{\pi/2}\frac{\pi}{2}(\sin(x)+\cos(x))^n\,dx = \frac{\pi}{4}\int_0^{\pi/2}\left ( \sqrt{2}\cdot cos(x-\frac{\pi}{4}) \right )^n\, dx && {\color{#D61F06} \sin(x)+\cos(x)=\sqrt{2}\cdot cos(x-\frac{\pi}{4}) }\\ & I_n= \frac{\pi}{4} \left ( \sqrt{2} \right )^n \int_{-\pi/4}^{\pi/4} \cos^n(u)\,du=\frac{\pi}{4} \left ( \sqrt{2} \right )^n 2\int_0^{\pi/4} \cos^n(u)\,du && {\color{#D61F06} let \, u=x-\frac{\pi}{4} \, and \, \cos(x) \text{ is even function}} \\ &I_n=\frac{\pi}{2} \left ( \sqrt{2} \right )^n \left ( \left . \frac{1}{n}\cos^{n-1}(x)sin(x) \right |_0^{\pi/4} +\frac{n-1}{n} \int_0^{\pi/4} \cos^{n-2}(x)\, dx\right ) && {\color{#D61F06} \int \cos^n(x)=\frac{1}{n}\cos^{n-1}(x)sin(x) +\frac{n-1}{n} \int \cos^{n-2}(x)\, dx}\\ &I_n=\frac{\pi}{2} \left ( \sqrt{2} \right )^n\left ( \frac{1}{n} \left ( \frac{1}{\sqrt{2}} \right )^n +\frac{n-1}{n} \int_0^{\pi/4} \cos^{n-2}(x)\, dx \right ) \\ &I_n=\frac{\frac{\pi}{2} + 2(n-1){\color{#3D99F6} \frac{\pi}{4} \left ( \sqrt{2} \right )^{n-2} 2\int_0^{\pi/4} \cos^{n-2}(x)\,du}}{n}\\ &I_n=\frac{\frac{\pi}{2}+2(n-1){\color{#3D99F6} I_{n-2}}}{n}\\ &126I_{126}= 126 \cdot \frac{\frac{\pi}{2}+250I_{124}}{126}=\frac{\pi}{2}+250I_{124}\\ &\frac{{\color{#3D99F6}126I_{126}}-\frac{\pi}{2}}{I_{124}}=\frac{{\color{#3D99F6}\frac{\pi}{2}+250I_{124}}-\frac{\pi}{2}}{I_{124}}=250 \end{aligned}$

[This solution is being edited.]