Kako integrirate int 1 / sqrt (x ^ 2-4x + 13) dx z uporabo trigonometrične substitucije?

Odgovor:

#int 1 / sqrt (x ^ 2-4x + 13) = l n | sqrt (1+ (x-2) ^ 2/9) + (x-2) / 3 | + C #

Pojasnilo:

#int 1 / sqrt (x ^ 2-4x + 13) d x = int 1 / sqrt (x ^ 2-4x + 9 + 4) d x #

#int 1 / (sqrt ((x-2) ^ 2 + 3 ^ 2)) d x #

# x-2 = 3tan theta "" d x = 3sec ^ 2 theta d theta #

#int 1 / sqrt (x ^ 2-4x + 13) dx = int (3sec ^ 2 theta d theta) / sqrt (9tan ^ 2 theta + 9) = int (3sec ^ 2 theta d theta) / (3sqrt (1 + tan ^ 2 theta)) "" 1 + tan ^ 2 theta = sek ^ 2 theta #

#int 1 / sqrt (x ^ 2-4x + 13) d x = int (3sec ^ 2 theta d theta) / (3sqrt (sec ^ 2 theta)) #

#int 1 / sqrt (x ^ 2-4x + 13) d x = int (prekliči (3sec ^ 2 theta) d theta) / (prekliči (3sec theta)) #

#int 1 / sqrt (x ^ 2-4x + 13) d x = int sek theta d theta #

#int 1 / sqrt (x ^ 2-4x + 13) d x = l n | sek theta + tan theta | + C #

#tan theta = (x-2) / 3 "" sek theta = sqrt (1 + tan ^ 2 theta) = sqrt (1+ (x-2) ^ 2/9) #

#int 1 / sqrt (x ^ 2-4x + 13) = l n | sqrt (1+ (x-2) ^ 2/9) + (x-2) / 3 | + C #

Odgovor:

# sinh ^ -1 ((x-2) / 3) + C #

Pojasnilo:

Možna je tudi hiperbolična različica:

• # x-2 = 3 sinh u #
• #dx = 3 cosh u du #

#int 1 / sqrt (x ^ 2-4x + 13) dx = int 1 / sqrt (9sinh ^ 2 u + 9) 3cosh u du = int 1 / (3cosh u) 3cosh u du = u + C #

Zato:

#int 1 / sqrt (x ^ 2-4x + 13) dx = sinh ^ -1 ((x-2) / 3) + C #