int_0^(pi/2) dx/(1+tanx) এর মান নির্ণয় করঃ

Explanation:

Another Explanation (5): সমাধান: ধরি, \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan x} \) 🤩 \( I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \frac{\sin x}{\cos x}} = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx \) ... (1) আমরা জানি, \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \) 😎 অতএব, \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos (\frac{\pi}{2} - x)}{\cos (\frac{\pi}{2} - x) + \sin (\frac{\pi}{2} - x)} dx \) \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx \) ... (2) এখন, (1) এবং (2) যোগ করে পাই, \( 2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx + \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx \) \( 2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x + \sin x}{\cos x + \sin x} dx = \int_{0}^{\frac{\pi}{2}} 1 dx \) 🙌 \( 2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \) সুতরাং, \( I = \frac{\pi}{4} \) 🎉 অতএব, \(\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \tan x} = \frac{\pi}{4}\) 👍