If f(x) is an odd function of x, then int_{ - | vedclass

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(C) Let $I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $. Since $\cos(-x) = \cos x$,the function $g(x) = f(\cos x)$ is an even functio

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$2\int_0^{\frac{\pi }{2}} {f(\sin x)\,dx} $

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(C) Let $I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $. Since $\cos(-x) = \cos x$,the function $g(x) = f(\cos x)$ is an even function because $g(-x) = f(\cos(-x)) = f(\cos x) = g(x)$. For an even function,the property $\int_{-a}^{a} g(x) \, dx = 2 \int_{0}^{a} g(x) \, dx$ holds. Thus,$I = 2 \int_{0}^{\frac{\pi }{2}} f(\cos x) \, dx$. Using the property $\int_{0}^{a} h(x) \, dx = \int_{0}^{a} h(a-x) \, dx$,we substitute $x = \frac{\pi}{2} - t$: $I = 2 \int_{0}^{\frac{\pi }{2}} f(\cos(\frac{\pi}{2} - t)) \, dt = 2 \int_{0}^{\frac{\pi }{2}} f(\sin t) \, dt$. Therefore,the correct option is $C$.

If $f(x)$ is an odd function of $x,$ then $\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $ is equal to

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