int_{0}^{pi / 2} frac{sin x-cos x}{1-sin x cd | vedclass

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(A) Let $I = \int_{0}^{\pi / 2} \frac{\sin x - \cos x}{1 - \sin x \cos x} d x$  $(i)$ Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) d

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(A) Let $I = \int_{0}^{\pi / 2} \frac{\sin x - \cos x}{1 - \sin x \cos x} d x$  $(i)$ Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $x$ with $(\frac{\pi}{2} - x)$: $I = \int_{0}^{\pi / 2} \frac{\sin(\frac{\pi}{2} - x) - \cos(\frac{\pi}{2} - x)}{1 - \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)} dx$ Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we get: $I = \int_{0}^{\pi / 2} \frac{\cos x - \sin x}{1 - \cos x \sin x} dx$ $I = - \int_{0}^{\pi / 2} \frac{\sin x - \cos x}{1 - \sin x \cos x} dx$ $I = -I$  (ii) Adding equation $(i)$ and (ii): $I + I = 0$ $2I = 0$ $I = 0$

$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x$ is equal to

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