Предмет: Математика,
автор: amberheart
Найти первообразные функций (их три, фото внутри)
Приложения:
Ответы
Автор ответа:
0
ж) sin(3x)/3 - ctg(x) + C
з)
![6x - 1 = t^2 \
t = sqrt{6x-1} \
dx = frac{1}{3} tdt \
frac{1}{3} intlimits^._ .{ frac{t^3 - t}{t+1} } , dt = frac{1}{3} intlimits^._ .{ t^2 - t} } , dt = frac{t^3}{9} - frac{t^2}{2} + C \
frac{(6x-1)^ frac{3}{2} }{9} - x + frac{1}{6} + C](https://tex.z-dn.net/?f=6x+-+1+%3D+t%5E2+%5C+%0At+%3D++sqrt%7B6x-1%7D++%5C+%0Adx+%3D++frac%7B1%7D%7B3%7D+tdt+%5C+%0A+frac%7B1%7D%7B3%7D++intlimits%5E._+.%7B+frac%7Bt%5E3+-+t%7D%7Bt%2B1%7D+%7D+%2C+dt+%3D+++frac%7B1%7D%7B3%7D++intlimits%5E._+.%7B+t%5E2+-+t%7D+%7D+%2C+dt+%3D+++frac%7Bt%5E3%7D%7B9%7D+-++frac%7Bt%5E2%7D%7B2%7D+%2B+C+%5C+%0A+frac%7B%286x-1%29%5E+frac%7B3%7D%7B2%7D+%7D%7B9%7D+-+x+%2B++frac%7B1%7D%7B6%7D+%2B+C "6x - 1 = t^2 \
t = sqrt{6x-1} \
dx = frac{1}{3} tdt \
frac{1}{3} intlimits^._ .{ frac{t^3 - t}{t+1} } , dt = frac{1}{3} intlimits^._ .{ t^2 - t} } , dt = frac{t^3}{9} - frac{t^2}{2} + C \
frac{(6x-1)^ frac{3}{2} }{9} - x + frac{1}{6} + C")
и)
![sin(x)*cos(x)*cos(2x)*cos(4x) = sin(8x)/8 \
frac{1}{8} intlimits^._. {sin(8x)} , dx = -frac{cos(8x)}{64} + C](https://tex.z-dn.net/?f=sin%28x%29%2Acos%28x%29%2Acos%282x%29%2Acos%284x%29+%3D+sin%288x%29%2F8+%5C+%0A+frac%7B1%7D%7B8%7D++intlimits%5E._.+%7Bsin%288x%29%7D+%2C+dx+%3D++-frac%7Bcos%288x%29%7D%7B64%7D+%2B+C+ "sin(x)*cos(x)*cos(2x)*cos(4x) = sin(8x)/8 \
frac{1}{8} intlimits^._. {sin(8x)} , dx = -frac{cos(8x)}{64} + C")
з)
и)
Похожие вопросы