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$\int_0^\pi \frac{\cos x}{\sqrt{1-\sin ^2 x}} d x=$
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$0$
Let $I=\int_0^\pi \frac{\cos x}{\sqrt{1-\sin ^2 x}} d x$
$\begin{aligned} & \Rightarrow I=\int_0^\pi \frac{\cos x}{|\cos x|} d x=\int_0^{\pi / 2} \frac{\cos x}{\cos x} d x+\int_{\frac{\pi}{2}}^\pi \frac{\cos x}{-\cos x} d x \\ & =\int_0^{\pi / 2} 1 d x-\int_{\pi / 2}^\pi 1 d x=\frac{\pi}{2}-\frac{\pi}{2}=0\end{aligned}$
$\begin{aligned} & \Rightarrow I=\int_0^\pi \frac{\cos x}{|\cos x|} d x=\int_0^{\pi / 2} \frac{\cos x}{\cos x} d x+\int_{\frac{\pi}{2}}^\pi \frac{\cos x}{-\cos x} d x \\ & =\int_0^{\pi / 2} 1 d x-\int_{\pi / 2}^\pi 1 d x=\frac{\pi}{2}-\frac{\pi}{2}=0\end{aligned}$
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