int_{0}^{frac{pi}{2}} frac{1-cot x}{operatorn | vedclass

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Question

(A) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x$.Converting to sine and cosine:$I = \int_{0}^{\frac{\pi}{2}}

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(A) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x$.Converting to sine and cosine:$I = \int_{0}^{\frac{\pi}{2}} \frac{1-\frac{\cos x}{\sin x}}{\frac{1}{\sin x}+\cos x} d x = \int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x \quad ...(1)$Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$:$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)} d x$$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x \sin x} d x \quad ...(2)$Adding equations $(1)$ and $(2)$:$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x + \cos x-\sin x}{1+\sin x \cos x} d x$$2I = \int_{0}^{\frac{\pi}{2}} 0 d x = 0$Therefore,$I = 0$.

$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$

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