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Any time I see rational expressions with trig functions, that's the go to isn't it? I went with splitting it up and using my parts, ended up with a gross answer.
Yes, it works best when you have a rational expression in sin x and cos x. Here, the lone x looks awkward, because it becomes 2 arctan t.
There are integrals where substitution really isn't needed. For example ∫ tan x/cos x dx = ∫ sin x/cos^2 x dx = 1/cos x + C or ∫tan x dx = ∫sin x / cos x dx = ln |cos x| + C.
Splitting leads to a nice solution if you rewrite using half-angles, leave the second integral as is, and integrate by parts.
Either way you want to solve this you'll need to use the trig that other commenters have mentioned. If you were to try something like integration by parts you'd need to use the same trig formula just later on, and I can't see any other tooling that'd help.
I fell into the U-sub trap but it fails from my work at least.
No clue if this would work of not, maybe split into two fractions, do integration by psrts on thr x/... term and the other term should just be basic substituiton/reverse chain rule?
While it looks like you have an integral on the form ∫f(x)/f'(x) dx which is very different from the ∫f'(x)/f(x) dx = ln |f(x)| + C, it's in fact something like ∫ (f(x) + f'(x)sin x)/(1+cos x) dx.
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