Co je f (x) = int -cos6x -3tanx dx pokud f (pi) = - 1?

Odpovědět:

Odpověď je:

#f (x) = - 1 / 6sin (6x) + 3ln | cosx | -1 #

Vysvětlení:

#f (x) = int (-cos6x-3tanx) dx #

#f (x) = - intcos (6x) dx-3inttanxdx #

Pro první integrál:

# 6x = u #

# (d (6x)) / (dx) = (du) / dx #

# 6 = (du) / dx #

# dx = (du) / 6 #

Proto:

#f (x) = - intcosu (du) / 6-3intsinx / cosxdx #

#f (x) = - 1 / 6intcosudu-3int ((- cosx) ') / cosxdx #

#f (x) = - 1 / 6intcosudu + 3int ((cosx) ') / cosxdx #

#f (x) = - 1 / 6sinu + 3ln | cosx | + c #

#f (x) = - 1 / 6sin (6x) + 3ln | cosx | + c #

Od té doby #f (π) = - 1 #

#f (π) = - 1 / 6sin (6π) + 3ln | cosπ | + c #

# -1 = -1 / 6 * 0 + 3ln | -1 | + c #

# -1 = 3ln1 + c #

# c = -1 #

Proto:

#f (x) = - 1 / 6sin (6x) + 3ln | cosx | -1 #