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, \( \int_0^{\pi/6} \sin 3x \cos 3x \, dx = \frac{1}{6} \) 🎉
