Question

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Answer 1

∫(3x²-(2/cos²x))dx=∫3x²dx-∫(2/cos²x)dx=3*∫x²dx-2*∫(1/cos²x)dx.
1. 3*∫x²dx=3*x³/3=x³=(π/3)³-(π/6)³=π³/27-π³/216=7π³/216.
2. 2*∫(1/cos²x)dx=2*tgx=2*(√3-√3/3)=2*(2√3/3)=4√3/3.
3. 7π³/216-4√3/3.
Ответ: 7π³/216-4√3/3.

Answer 2

$displaystyle intlimits^frac{pi}{6}_frac{pi}{3} (3x^2-frac{2}{cos^2x})dx=-intlimits^frac{pi}{3}_frac{pi}{6} (3x^2-frac{2}{cos^2x})dx=\=(2tgx-x^3)|^frac{pi}{3}_frac{pi}{6}=2sqrt3-frac{pi^3}{27}-frac{2}{sqrt3}+frac{pi^3}{216}approx1,30$