F (x) = int -cos6x -3tanx dx ถ้า f (pi) = - 1 คืออะไร

ตอบ:

คำตอบคือ:

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

คำอธิบาย:

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

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

สำหรับอินทิกรัลแรก:

# 6x u = #

# (ง (6x)) / (DX) = (du) / DX #

# 6 = (du) / DX #

# DX = (du) / 6 #

ดังนั้น:

# 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 #

ตั้งแต่ # f (π) = - 1 #

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

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

# -1 = 3ln1 + C #

# c = -1 #

ดังนั้น:

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