Question

write in a notebook! Help please

Answer 1

Ответ: $\sqrt3-\frac{\pi}{3}$

Пошаговое объяснение:

$\int\limits^\frac{\pi}{3}_0 {tg^2x} \, dx =\int\limits^\frac{\pi}{3}_0 {\frac{sin^2x}{cos^2x} } \, dx=\int\limits^\frac{\pi}{3}_0 {\frac{1-cos^2x}{cos^2x}} \, dx=\int\limits^\frac{\pi}{3}_0 {\frac{1}{cos^2x}} \, dx-\int\limits^\frac{\pi}{3}_0 {1} \, dx=\\\\=(tg\ x-x)|_0^\frac{\pi}{3}=tg\ \frac{\pi}{3}-\frac{\pi}{3}=\sqrt3-\frac{\pi}{3}$