Question

Вычислите интеграл
$s {3}^{6} (4x - \frac{1}{2 \sqrt{x - 2} } )dx - s{0}^{ \frac{\pi}{4} } \frac{dx}{ \cos^{2}x }$
​

Answer 1

Ответ:

$\displaystyle \int\limits^6_3\, \Big((4x-\frac{1}{2\sqrt{x-2}}\Big)\, dx-\int\limits_0^{\frac{\pi }{4}}\, \frac{dx}{cos^2x}=\Big(2x^2-\frac{1}{2}\cdot 2\sqrt{x-2}\Big)\Big|_3^6-tgx\, \Big|_0^{\frac{\pi}{4}}=\\\\\\=2\cdot 36-\sqrt{4}-2\cdot 9+\sqrt{1}=72-2-18+1=53$