Prove that

$$\int\_{0}^{\frac{\pi}{2}}(\sin x)^ndx = \int\_{0}^{\frac{\pi}{2}}(\cos x)^ndx.$$

My work.

Let's consider:
$x = \frac{\pi}{2}-x$

So

$$-\int\_{0}^{\frac{\pi}{2}}\left(\sin\left(\frac{\pi}{2}-x\right)\right)^{n}dx=- \int \_{0}^{\frac{\pi}{2}}(\cos x)^ndx.$$

What did wrong that I am getting with $-$?