Kako integrirate int dx / (x ^ 2 + 1) ^ 2 z uporabo substitucij trigonometrije?

Odgovor:

#int dx / (x ^ 2 + 1) ^ 2 = (1/2) (tan ^ -1 (x) + x / (1 + x ^ 2)) #

Pojasnilo:

#int dx / (x ^ 2 + 1) ^ 2 #

Uporaba # x = tan (a) #

# dx = sek ^ 2 (a) da #

# intdx / (x ^ 2 + 1) ^ 2 = int (sek ^ 2 (a) da) / (1 + tan ^ 2a) ^ 2 #

Uporabite identiteto # 1 + tan ^ 2 (a) = sek ^ 2 (a) #

# intdx / (x ^ 2 + 1) ^ 2 = int (sek ^ 2 (a) da) / sec ^ 4 (a) #

# = int (da) / sec ^ 2 (a) #

# = int cos ^ 2 (a) da #

# = int ((1 + cos (2a)) / 2) da #

# = (1/2) (int (da) + int cos (2a) da) #

# = (1/2) (a + sin (2a) / 2) #

# = (1/2) (a + (2sin (a) cos (a)) / 2) #

# = (1/2) (a + sin (a).cos (a)) #

to vemo # a = tan ^ -1 (x) #

#sin (a) = x / (sqrt (1 + x ^ 2) #

#cos (a) = x / (sqrt (1 + x ^ 2 #)

#int dx / (x ^ 2 + 1) ^ 2 = (1/2) (tan ^ -1 (x) + sin (sin ^ -1 (x / (sqrt (1 + x ^ 2))) cos (cos ^ -1 (1 / (sqrt (1 + x ^ 2)))) #

# = (1/2) (tan ^ -1 (x) + (x / (sqrt (1 + x ^ 2)) 1 / sqrt (1 + x ^ 2)) #

# = (1/2) (tan ^ -1 (x) + x / (1 + x ^ 2)) #

Odgovor:

#int dx / (x ^ 2 + 1) ^ 2 = 1/2 (arctan (x) + x / (x ^ 2 + 1)) #

Pojasnilo:

#int dx / (x ^ 2 + 1) ^ 2 # zamenjavo

#x = tan (y) # in posledično

#dx = dy / (cos (y) ^ 2) #

imamo

#int dx / (x ^ 2 + 1) ^ 2 equiv int dy / (cos (y) ^ 2 (1 / cos (y) ^ 4)) = int cos (y) ^ 2dy #

ampak

# d / (dy) (sin (y) cos (y)) = cos (y) ^ 2-sin (y) ^ 2 = 2 cos (y) ^ 2-1 #

potem

#int cos (y) ^ 2 dy = 1/2 (y + sin (y) cos (y)) #

Končno, spomnite se #y = arctan (x) # imamo

#int dx / (x ^ 2 + 1) ^ 2 = 1/2 (arctan (x) + x / (x ^ 2 + 1)) #