Pregunta # f9641

Responder:

#int cos (x) / (sin ^ 2 (x) + sin (x)) "d" x = ln | sin (x) / (sin (x) +1) | + C #

Explicación:

# int cos (x) / (sin ^ 2 (x) + sin (x)) "d" x #

Sustituir # u = pecado (x) # y # "d" u = cos (x) "d" x #. Esto da

# = int ("d" u) / (u ^ 2 + u) #

# = int ("d" u) / (u (u + 1)) #

Separar a fracciones parciales ya que # 1 / (u (u + 1)) = 1 / u-1 / (u + 1) #:

# = int (1 / u-1 / (u + 1)) "d" u #

# = ln | u | -ln | u + 1 | + C #

# = ln | u / (u + 1) | + C #

Sustituir la espalda # u = pecado (x) #:

# = ln | sin (x) / (sin (x) +1) | + C #