intsin^2 4xdx =কত?

Explanation:

Another Explanation (5): সমাধান: 🤔 আমরা জানি, \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \) তাহলে, \( \sin^2 4x = \frac{1 - \cos 8x}{2} \) সুতরাং, \( \int \sin^2 4x \, dx = \int \frac{1 - \cos 8x}{2} \, dx \) \( = \frac{1}{2} \int (1 - \cos 8x) \, dx \) \( = \frac{1}{2} \left[ \int 1 \, dx - \int \cos 8x \, dx \right] \) \( = \frac{1}{2} \left[ x - \frac{\sin 8x}{8} \right] + C \) \( = \frac{1}{2}x - \frac{1}{16}\sin 8x + C \) \( = \frac{1}{16}(8x - \sin 8x) + C \) 🥳 সুতরাং, \( \int \sin^2 4x \, dx = \frac{1}{16}(8x - \sin 8x) + C \) 🤩