Вычислить интеграл π\12|0 (sin3t+cost)^2dt

Решите задачу:

$\int\limits_0^{\frac{\pi}{12}} (sin3t+cost)^2dt= \int\limits^{\frac{\pi}{12}}_0(sin^23t+2sin3t\, cost+cos^2t)dt=\\\\=\int\limits^{\frac{\pi }{12}}_0\Big (\frac{1-cos6t}{2}dt+2\cdot (3sint-4sin^3t)cost+\frac{1+cos2t}{2}\Big )dt=\\\\= \int\limits^{\frac{\pi}{12}}_0\Big (\frac{1}{2}-\frac{1}{2}cos6t+6sint\cdot cost-8sin^3t\cdot cost+\frac{1}{2}+\frac{1}{2}cos2t\Big )dt=\\\\=(t-\frac{1}{2\cdot 6}sin6t+6\cdot \frac{sin^2t}{2}-8\cdot \frac{sin^4t}{4}+\frac{1}{2 \cdot 2}sin2t\Big )\Big |_0^{ \frac{\pi}{12}}=$

$=\frac{\pi }{12}-\frac{1}{12}+3\cdot sin^2\frac{\pi }{12}-2\cdot sin^4\frac{\pi }{12}+\frac{1}{4}=\\\\=\frac{1}{6}+\frac{\pi}{12}+3\cdot \Big (\frac{\sqrt6-\sqrt2}{4}\Big )^2-2\cdot \Big (\frac{\sqrt6-\sqrt2}{4}\Big )^4=\\\\=\frac{2+\pi }{12}+3\cdot \frac{8-2\sqrt{12}}{16}-2\cdot (\frac{8-2\sqrt{12}}{16})^2=\\\\=\frac{2+\pi }{12}+\frac{6-3\sqrt3}{4}-2\cdot \frac{112-64\sqrt3}{256}=\frac{2+\pi }{12}+\frac{3}{2}-\frac{3\sqrt3}{4}-\frac{7}{8}+\frac{\sqrt3}{2}=\\\\= \frac{\pi }{12}+\frac{19}{24}-\frac{\sqrt3}{4}=\frac{2\pi +19-6\sqrt3}{24}$