int_0^pi frac{cos x}{sqrt{1-sin ^2 x}} d x= | vedclass

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(D) Let $I = \int_0^\pi \frac{\cos x}{\sqrt{1-\sin^2 x}} dx$. Since $\sqrt{1-\sin^2 x} = |\cos x|$,the integral becomes $I = \int_0^\pi \frac{\cos x}{

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(D) Let $I = \int_0^\pi \frac{\cos x}{\sqrt{1-\sin^2 x}} dx$. Since $\sqrt{1-\sin^2 x} = |\cos x|$,the integral becomes $I = \int_0^\pi \frac{\cos x}{|\cos x|} dx$. We know that $\cos x > 0$ for $x \in [0, \pi/2)$ and $\cos x Thus,we split the integral at $x = \pi/2$: $I = \int_0^{\pi/2} \frac{\cos x}{\cos x} dx + \int_{\pi/2}^\pi \frac{\cos x}{-\cos x} dx$. $I = \int_0^{\pi/2} 1 dx - \int_{\pi/2}^\pi 1 dx$. $I = [x]_0^{\pi/2} - [x]_{\pi/2}^\pi$. $I = (\pi/2 - 0) - (\pi - \pi/2) = \pi/2 - \pi/2 = 0$.

$\int_0^\pi \frac{\cos x}{\sqrt{1-\sin ^2 x}} d x=$

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