int_0^(pi/4)(cos2theta)/cos^2 theta theta=? 

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \(cos2\theta = cos^2\theta - sin^2\theta\)। সুতরাং, \[ \int_{0}^{\frac{\pi}{4}} \frac{cos2\theta}{cos^2\theta} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{cos^2\theta - sin^2\theta}{cos^2\theta} d\theta \] \[ = \int_{0}^{\frac{\pi}{4}} (1 - \frac{sin^2\theta}{cos^2\theta}) d\theta = \int_{0}^{\frac{\pi}{4}} (1 - tan^2\theta) d\theta \] আমরা জানি, \(sec^2\theta - tan^2\theta = 1\), সুতরাং \(tan^2\theta = sec^2\theta - 1\)। \[ = \int_{0}^{\frac{\pi}{4}} (1 - (sec^2\theta - 1)) d\theta = \int_{0}^{\frac{\pi}{4}} (2 - sec^2\theta) d\theta \] \[ = [2\theta - tan\theta]_{0}^{\frac{\pi}{4}} = (2 \cdot \frac{\pi}{4} - tan(\frac{\pi}{4})) - (2 \cdot 0 - tan(0)) \] \[ = (\frac{\pi}{2} - 1) - (0 - 0) = \frac{\pi}{2} - 1 \] সুতরাং, \(\int_{0}^{\frac{\pi}{4}} \frac{cos2\theta}{cos^2\theta} d\theta = \frac{\pi}{2} - 1\) 🥳