How do you integrate #tanx / (secx + cosx)#?
1 Answer
Aug 1, 2016
Explanation:
We should first try to simplify the integrand.
#tanx/(secx+cosx)=(sinx/cosx)/(1/cosx+cosx)=sinx/(cosx(1/cosx+cosx))=sinx/(1+cos^2x)#
Thus:
#inttanx/(secx+cosx)dx=intsinx/(1+cos^2x)dx#
We can use substitution here. Let
#=-int(-sinx)/(1+cos^2x)dx=-int1/(1+u^2)du#
This is the arctangent integral!
#=-arctan(u)+C=-arctan(cosx)+C#