Thread: A nice integral

It looks very nice but ... it's quite simple

$\displaystyle \int_0^{\pi} \frac{ dx}{ 1 + \sin^{\cos(x)}(x)} $

$\displaystyle I=\int_0^{\pi}\frac{dx}{1+(\sin x)^{\cos x}}$

Let $\displaystyle x=\pi-t\Rightarrow dx=-dt$

$\displaystyle x=0\Rightarrow t=\pi, \ x=\pi\Rightarrow t=0$

Then $\displaystyle I=-\int_{\pi}^0\frac{dt}{1+(\sin(\pi-t))^{\cos(\pi-t)}}=\int_0^{\pi}\frac{dt}{1+(\sin t)^{-\cos t}}=$

$\displaystyle =\int_0^{\pi}\frac{(\sin t)^{\cos t}}{1+(\sin t)^{\cos t}}dt=\int_0^{\pi}\left(1-\frac{1}{1+(\sin t)^{\cos t}}\right)dt=$

$\displaystyle =\left. t\right|_0^{\pi}-I=\pi-I\Rightarrow I=\frac{\pi}{2}$