How do I evaluate #int(1-sinx)/cosx dx#?

1 Answer
Jan 31, 2015

#=ln(1+sin(x))+c#

Well, this one is a little bit tricky...
I started with the idea that the result must be a logarithm...
So I did a little manipulation to get to a friendlier version of your function by multiplying and dividing by:

#(1+sin(x))/(1+sin(x))#; so I get:

#int(1-sin(x))/cos(x)*(1+sin(x))/(1+sin(x))dx=#
#=int(1-sin^2(x))/(cos(x)(1+sin(x)))dx=#
#=int(cos^2(x))/(cos(x)(1+sin(x)))dx=#
#=int(cos(x))/((1+sin(x)))dx=# which is a manipulated version of your original function and we can call it (1). The integral of (1) is indeed a logarithm:

#=ln(1+sin(x))+c#

Which derived gives (1) or your original function!!!!!