A nice integral

**simplependulum** · August 11th 2009, 10:43 PM

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

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

**red_dog** · August 12th 2009, 12:12 AM

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

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

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

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

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

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

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