Question

Вычислить определенный интеграл. Задание на фото

Answer 1

$1)\ \ \int\limits^6_3\, \dfrac{\sqrt{x^2-9}}{x^4}\, dx=\Big[\ x=\dfrac{3}{cost}\ ,\ dx=\dfrac{3\, sint\, dt}{cos^2t}\ ,\ x^2-9=\dfrac{9}{cos^2t}-9=9\, tg^2t\ , \\\\\\t_1=arccos1=0\ ,\ t_2=arccos\dfrac{1}{2}=\dfrac{\pi}{3}\ \Big]=\int\limits^{\pi /3}_0\, \dfrac{3\, tgt}{\dfrac{3^4}{cos^4t}}\cdot \dfrac{3\, sint\, dt}{cos^2t}=$

$=\dfrac{1}{9}\int\limits^{\pi /3}_0\, sin^2t\cdot cost\, dt=\dfrac{1}{9}\cdot \dfrac{sin^3t}{3}\, \Big|_0^{\pi /3}=\dfrac{1}{27}\cdot \Big(\dfrac{\sqrt3}{2}\, \Big)^3=\dfrac{\sqrt3}{72}$

$2)\ \ \int\limits^{\pi }_0\, sin^4\dfrac{x}{2}\, dx=\int\limits^{\pi }_0\, \Big(\dfrac{1-cosx}{2}\Big)^2\, dx=\dfrac{1}{4}\int\limits^{\pi }_0\, \Big(1-2cosx+cos^2x\Big)\, dx=\\\\\\=\dfrac{1}{4}\int\limits^{\pi }_0\, \Big(1-2cosx+\dfrac{1+cos2x}{2}\Big)\, dx=\dfrac{1}{4}\int\limits^{\pi }_0\, \Big(\dfrac{3}{2}-2cosx+\dfrac{cos2x}{2}\Big)\, dx=\\\\\\=\dfrac{1}{4}\cdot \Big(\dfrac{3x}{2}-2sinx+\dfrac{1}{4}\cdot sin2x\Big)\Big|_0^{\pi }=\dfrac{1}{4}\cdot \dfrac{3\pi }{2}=\dfrac{3\pi }{8}$