מה זה f (x) = int-cos6x -3tanx dx אם f (pi) = 1?

תשובה:

תשובה היא:

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

הסבר:

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

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

עבור האינטגרל הראשון:

# 6x = u #

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

# 6 = (du) / dx #

# dx = (du) / 6 #

לכן:

#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 #

מאז #f (π) = - 1 #

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

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

# -1 = 3ln1 + c #

# c = -1 #

לכן:

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