int_0^(pi/2)1/(1+cosx)dx= কত?

প্রশ্ন: \( \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos x} dx = ? \)

সমাধান:

আমরা জানি, \( \cos x = \frac{1 - \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})} \)

সুতরাং, \( \frac{1}{1 + \cos x} = \frac{1}{1 + \frac{1 - \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})}} = \frac{1 + \tan^2(\frac{x}{2})}{1 + \tan^2(\frac{x}{2}) + 1 - \tan^2(\frac{x}{2})} = \frac{1 + \tan^2(\frac{x}{2})}{2} = \frac{\sec^2(\frac{x}{2})}{2} \)

এখন, \( \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos x} dx = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2(\frac{x}{2})}{2} dx \)

ধরি, \( u = \frac{x}{2} \), তাহলে \( du = \frac{1}{2} dx \) বা \( dx = 2 du \)

যখন \( x = 0 \), \( u = 0 \) এবং যখন \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{4} \)

সুতরাং, \( \int_{0}^{\frac{\pi}{2}} \frac{\sec^2(\frac{x}{2})}{2} dx = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2(u)}{2} (2 du) = \int_{0}^{\frac{\pi}{4}} \sec^2(u) du \)

আমরা জানি, \( \int \sec^2(u) du = \tan(u) + C \)

অতএব, \( \int_{0}^{\frac{\pi}{4}} \sec^2(u) du = [\tan(u)]_{0}^{\frac{\pi}{4}} = \tan(\frac{\pi}{4}) - \tan(0) = 1 - 0 = 1 \)

সুতরাং, \( \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos x} dx = 1 \) 🎉