int_(0)^(π/2)dx/(1+cosx) এর মান কত ?

Explanation:

Another Explanation (5): সমাধান: আমরা \( \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos x} \) এর মান নির্ণয় করব। আমরা জানি, \( \cos x = \frac{1 - \tan^2{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}} \) সুতরাং, \( \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos x} = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \frac{1 - \tan^2{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}}} \) \( = \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}} + 1 - \tan^2{\frac{x}{2}}} dx \) \( = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2{\frac{x}{2}}}{2} dx \) \( = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sec^2{\frac{x}{2}} dx \) এখন, \( \tan{\frac{x}{2}} \) এর অন্তরকলজ \( \frac{1}{2} \sec^2{\frac{x}{2}} \) সুতরাং, \( = \left[ \tan{\frac{x}{2}} \right]_{0}^{\frac{\pi}{2}} \) \( = \tan{\frac{\pi}{4}} - \tan{0} \) \( = 1 - 0 \) \( = 1 \) অতএব, \( \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos x} = 1 \) সুতরাং, উত্তর: 1 🎉