What is the integral of #int (1-sinx)/(cosx) dx#?
1 Answer
Feb 24, 2016
# = (1-sin^2x)/(cosx(1+sinx))#
# = cos^2x/(cosx(1+sinx))#
# = cosx/(1+sinx)#
# = ln(1+sinx)+C# (by substitution with
#u=1+sinx# )
