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

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(C) Let $ I = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 - \sin x \cos x} dx $  $(1)$  Using the property $ \int_0^a f(x) dx = \int_0^a f(a-x) dx

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(C) Let $ I = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 - \sin x \cos x} dx $  $(1)$  Using the property $ \int_0^a f(x) dx = \int_0^a f(a-x) dx $,we get:  $ I = \int_0^{\frac{\pi}{2}} \frac{\sin(\frac{\pi}{2} - x) - \cos(\frac{\pi}{2} - x)}{1 - \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)} dx $  $ I = \int_0^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 - \cos x \sin x} dx $  $(2)$  Adding equation $(1)$ and $(2)$:  $ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{\sin x - \cos x}{1 - \sin x \cos x} + \frac{\cos x - \sin x}{1 - \sin x \cos x} \right) dx $  $ 2I = \int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x + \cos x - \sin x}{1 - \sin x \cos x} dx $  $ 2I = \int_0^{\frac{\pi}{2}} 0 dx = 0 $  Therefore,$ I = 0 $.

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

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