$$I = \int\sin x \log_e(\sec x + \tan x) dx$$
My attempt: $$\int \sin x \log_e(\sec x + \tan x) dx = \int\sin x\log_e(\frac{\sin x+1}{\cos x}) dx$$ $$\therefore I = \int \sin x\log_e(\sin x+1) dx - \int \sin x \log_e(\cos x) dx$$
I can solve the second integral but how do I go about solving the first one.
You can use integration by parts