int_0^(π/4)(cos2x)/cos^2xdx =কত?

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \( \cos 2x = \cos^2 x - \sin^2 x \) সুতরাং, \[ \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{\cos^2 x} dx = \int_0^{\frac{\pi}{4}} \frac{\cos^2 x - \sin^2 x}{\cos^2 x} dx \] \[ = \int_0^{\frac{\pi}{4}} \left(1 - \frac{\sin^2 x}{\cos^2 x}\right) dx = \int_0^{\frac{\pi}{4}} (1 - \tan^2 x) dx \] আমরা জানি, \( \sec^2 x - \tan^2 x = 1 \) , সুতরাং \( \tan^2 x = \sec^2 x - 1 \) \[ \int_0^{\frac{\pi}{4}} (1 - \tan^2 x) dx = \int_0^{\frac{\pi}{4}} (1 - (\sec^2 x - 1)) dx \] \[ = \int_0^{\frac{\pi}{4}} (2 - \sec^2 x) dx = \left[2x - \tan x\right]_0^{\frac{\pi}{4}} \] \[ = \left(2 \cdot \frac{\pi}{4} - \tan \frac{\pi}{4}\right) - \left(2 \cdot 0 - \tan 0\right) \] \[ = \left(\frac{\pi}{2} - 1\right) - (0 - 0) = \frac{\pi}{2} - 1 \] অতএব, \( \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{\cos^2 x} dx = \frac{\pi}{2} - 1 \) 🥳