int_0^(pi/2) (dx)/(1+cosx) এর মান কত? 

Explanation:

Another Explanation (5): ইন্টিগ্রেশনটির মান নির্ণয়: \[ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\cos x} \] আমরা জানি, \(\cos x = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\) সুতরাং, \[ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}} \] \[ = \int_{0}^{\frac{\pi}{2}} \frac{1+\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}+1-\tan^2\frac{x}{2}} dx \] \[ = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2\frac{x}{2}}{2} dx \] \[ = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sec^2\frac{x}{2} dx \] এখন, \(\frac{x}{2} = z\) ধরলে, \(\frac{1}{2}dx = dz\) বা \(dx = 2dz\) যখন \(x = 0\), তখন \(z = 0\) এবং যখন \(x = \frac{\pi}{2}\), তখন \(z = \frac{\pi}{4}\) সুতরাং, \[ = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \sec^2 z \cdot 2 dz \] \[ = \int_{0}^{\frac{\pi}{4}} \sec^2 z dz \] \[ = [\tan z]_{0}^{\frac{\pi}{4}} \] \[ = \tan \frac{\pi}{4} - \tan 0 \] \[ = 1 - 0 = 1 \] অতএব, \[ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\cos x} = 1 \] 🥳🥳🥳