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

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

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

$\int_0^{\pi /2} \frac{\cos x - \sin x}{1 + \sin x \cos x} \,dx = $

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