intsin^2 4xdx=? 

Explanation:

Another Explanation (5): সমাধান: \( \int \sin^2(4x) \, dx \) নির্ণয় করতে হবে। আমরা জানি, \( \cos(2\theta) = 1 - 2\sin^2(\theta) \) সুতরাং, \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \) অতএব, \( \sin^2(4x) = \frac{1 - \cos(8x)}{2} \) 😃 তাহলে, \(\begin{aligned} \int \sin^2(4x) \, dx &= \int \frac{1 - \cos(8x)}{2} \, dx \\ &= \frac{1}{2} \int (1 - \cos(8x)) \, dx \\ &= \frac{1}{2} \left[ \int 1 \, dx - \int \cos(8x) \, dx \right] \\ &= \frac{1}{2} \left[ x - \frac{\sin(8x)}{8} \right] + C \\ &= \frac{1}{2}x - \frac{1}{16}\sin(8x) + C \\ &= \frac{1}{16} (8x - \sin(8x)) + C \end{aligned}\) ✨ সুতরাং, \( \int \sin^2(4x) \, dx = \frac{1}{16}(8x - \sin(8x)) + C \) 🎉