int_0^(π/6)sin3xcos3xdx =?

Explanation:

Another Explanation (5): সমাধান: 🤔 আমরা জানি, \( \sin 2\theta = 2 \sin \theta \cos \theta \). সুতরাং, \( \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \) অতএব, \( \sin 3x \cos 3x = \frac{1}{2} \sin 6x \) এখন, \( \int_0^{\pi/6} \sin 3x \cos 3x \, dx = \int_0^{\pi/6} \frac{1}{2} \sin 6x \, dx \) \( = \frac{1}{2} \int_0^{\pi/6} \sin 6x \, dx \) \( = \frac{1}{2} \left[ -\frac{1}{6} \cos 6x \right]_0^{\pi/6} \) \( = -\frac{1}{12} \left[ \cos 6x \right]_0^{\pi/6} \) \( = -\frac{1}{12} \left[ \cos \left(6 \cdot \frac{\pi}{6}\right) - \cos (6 \cdot 0) \right] \) \( = -\frac{1}{12} \left[ \cos \pi - \cos 0 \right] \) \( = -\frac{1}{12} \left[ -1 - 1 \right] \) \( = -\frac{1}{12} (-2) \) \( = \frac{2}{12} \) \( = \frac{1}{6} \) সুতরাং, \( \int_0^{\pi/6} \sin 3x \cos 3x \, dx = \frac{1}{6} \) 🎉