int sin^2 4xdx =কত?

Explanation:

Another Explanation (5): সমাধান: আমরা জানি, \(sin^2x = \frac{1 - cos2x}{2}\) সুতরাং, \(sin^2 4x = \frac{1 - cos8x}{2}\) এখন, \(\int sin^2 4x dx = \int \frac{1 - cos8x}{2} dx\) \(= \frac{1}{2} \int (1 - cos8x) dx\) \(= \frac{1}{2} [\int 1 dx - \int cos8x dx]\) \(= \frac{1}{2} [x - \frac{sin8x}{8}] + c\) \(= \frac{1}{2}x - \frac{1}{16}sin8x + c\) \(= \frac{1}{16}(8x - sin8x) + c\) 🎉 অতএব, \(\int sin^2 4x dx = \frac{1}{16}(8x - sin8x) + c\) 🥳