How do you integrate #1/(1+tanx) dx#?
3 Answers
Use the substitution
Explanation:
Let
#I=int1/(1+tanx)dx#
Apply the substitution
#I=int1/((1+u^2)(1+u))du#
Apply partial fraction decomposition:
#I=1/2int((1-u)/(1+u^2)+1/(1+u))du#
Rearrange:
#I=1/2int(1/(1+u^2)-1/2(2u)/(1+u^2)+1/(1+u))du#
Integrate term by term:
#I=1/2(tan^(-1)u-1/2ln(1+u^2)+ln(1+u))+C#
Reverse the substitution:
#I=1/2(x-ln(secx)+ln(1+tanx))+C#
Simplify:
#I=1/2(x+ln(sinx+cosx))+C#
Explanation:
We want to find
We start by substituting
So
Now we perform a partial fraction decomposition on the integrand
Letting
Letting
Letting
So,
and
Reversing for
Explanation:
Let,