How do you integrate #tanx / (secx + cosx)#?

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Answer 1 · 2016-08-01T05:07:17

#-arctan(cosx)+C#

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 #u=cosx#. This implies that #du=-sinxdx#.

#=-int(-sinx)/(1+cos^2x)dx=-int1/(1+u^2)du#

This is the arctangent integral!

#=-arctan(u)+C=-arctan(cosx)+C#