int_{0}^{pi / 2} frac{d x}{1+tan x} is equal | vedclass

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(D) Let $I = \int_{0}^{\pi / 2} \frac{d x}{1+	an x}$.We can write $	an x = \frac{\sin x}{\cos x}$,so $I = \int_{0}^{\pi / 2} \frac{\cos x}{\sin x +

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$\pi / 4$

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(D) Let $I = \int_{0}^{\pi / 2} \frac{d x}{1+	an x}$.We can write $	an x = \frac{\sin x}{\cos x}$,so $I = \int_{0}^{\pi / 2} \frac{\cos x}{\sin x + \cos x} d x$ ... $(i)$.Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:$I = \int_{0}^{\pi / 2} \frac{\cos(\pi / 2 - x)}{\sin(\pi / 2 - x) + \cos(\pi / 2 - x)} d x = \int_{0}^{\pi / 2} \frac{\sin x}{\cos x + \sin x} d x$ ... $(ii)$.Adding $(i)$ and $(ii)$:$2I = \int_{0}^{\pi / 2} \frac{\cos x + \sin x}{\sin x + \cos x} d x = \int_{0}^{\pi / 2} 1 d x$.$2I = [x]_{0}^{\pi / 2} = \pi / 2$.Therefore,$I = \pi / 4$.

$\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$ is equal to

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