How do I evaluate #int(1 + cosx)/sinx d x#?

1 Answer
Jan 31, 2015

#=-ln(1+cos(x))+c#

Well, this one is another one 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+cos(x))/(1+cos(x))#; so I get:

#int(1-cos(x))/sin(x)*(1+cos(x))/(1+cos(x))dx=#
#=int(1-cos^2(x))/(sin(x)(1+cos(x)))dx=#
#=int(sin^2(x))/(sin(x)(1+sin(x)))dx=#
#=int(sin(x))/((1+cos(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+cos(x))+c#

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