By using the properties of definite integrals | vedclass

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(A) 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}

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(A) 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:$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin(\frac{\pi}{2} - x) - \cos(\frac{\pi}{2} - x)}{1 + \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)} dx$Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we have:$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 + \cos x \sin x} dx$ ..... $(2)$Adding equations $(1)$ and $(2)$:$I + I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx + \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 + \sin x \cos x} 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}} \frac{0}{1 + \sin x \cos x} dx$$2I = 0$$I = 0$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$.

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