Mikä on f (x) = int -cos6x -3tanx dx, jos f (pi) = - 1?

Vastaus:

Vastaus on:

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

Selitys:

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

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

Ensimmäinen integraali:

# 6x = u #

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

# 6 = (du) / dx #

# Dx = (du) / 6 #

Siksi:

#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 #

Siitä asti kun #f (π) = - 1 #

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

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

# -1 = 3ln1 + c #

# C = -1 #

Siksi:

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