int_o^(pi/2)dx/(1+cosx)=? 

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos x}\) 🤓 আমরা জানি, \(\cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\) 😎 সুতরাং, \(I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}\) ✨ \(I = \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} + 1 - \tan^2 \frac{x}{2}} dx\) 🤩 \(I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 \frac{x}{2}}{2} dx\) 🥳 \(I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sec^2 \frac{x}{2} dx\) 🤯 এখন, \(\frac{x}{2} = z\) ধরি। তাহলে, \(\frac{1}{2} dx = dz\) বা \(dx = 2 dz\) হবে। 🧐 যখন \(x = 0\), তখন \(z = 0\)। 🤓 যখন \(x = \frac{\pi}{2}\), তখন \(z = \frac{\pi}{4}\)। 😎 সুতরাং, \(I = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \sec^2 z \cdot 2 dz\) 😶‍🌫️ \(I = \int_{0}^{\frac{\pi}{4}} \sec^2 z dz\) 😵‍💫 \(I = [\tan z]_{0}^{\frac{\pi}{4}}\) 🤠 \(I = \tan \frac{\pi}{4} - \tan 0\) 😈 \(I = 1 - 0\) 🤡 \(I = 1\) 🥰 অতএব, \(\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos x} = 1\) 😇