int_0^(π/2) sin^2 2θdθ =?

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \( \sin^2 x = \frac{1 - \cos 2x}{2} \)। সুতরাং, \[ \sin^2 2\theta = \frac{1 - \cos 4\theta}{2} \] এখন, \[ \int_0^{\pi/2} \sin^2 2\theta \, d\theta = \int_0^{\pi/2} \frac{1 - \cos 4\theta}{2} \, d\theta \] \[ = \frac{1}{2} \int_0^{\pi/2} (1 - \cos 4\theta) \, d\theta \] \[ = \frac{1}{2} \left[ \int_0^{\pi/2} 1 \, d\theta - \int_0^{\pi/2} \cos 4\theta \, d\theta \right] \] \[ = \frac{1}{2} \left[ \theta \Big|_0^{\pi/2} - \frac{\sin 4\theta}{4} \Big|_0^{\pi/2} \right] \] \[ = \frac{1}{2} \left[ \frac{\pi}{2} - 0 - \frac{1}{4} (\sin 2\pi - \sin 0) \right] \] \[ = \frac{1}{2} \left[ \frac{\pi}{2} - \frac{1}{4} (0 - 0) \right] \] \[ = \frac{1}{2} \cdot \frac{\pi}{2} \] \[ = \frac{\pi}{4} \] অতএব, \( \int_0^{\pi/2} \sin^2 2\theta \, d\theta = \frac{\pi}{4} \)। 🎉