integrate 1/(5+sinx-3cosx)

let u = tan(x/2)
cosx = (1-u^2)/(1+u^2)
sinx = 2u/(1+u^2)
dx = 2/(1+u^2)

Question

integrate 1/(5+sinx-3cosx)

let u = tan(x/2)
cosx = (1-u^2)/(1+u^2)
sinx = 2u/(1+u^2)
dx = 2/(1+u^2)

anonymous · Answer

Oh my god... This is a monster to integrate...

anonymous · Answer

yeah, after substituting everything i got up to
1/[4(u+(1/8))^2 + (15/64)]

anonymous · Answer

if you could help me with that, that'd be cool. I don't know what to do next to integrate

anonymous · Answer

You don't have to use intergration by parts. You can use substitution twice.

anonymous · Answer

?

anonymous · Answer

Nevermind. I am wrong.

anonymous · Answer

Dammit. Sorry. I can't solve this :( . It's beyond my level of math.

anonymous · Answer

quick unrelated question: if i have 1/u(1-u^2) can i use partial fractions to integrate? even though its (-u^2 + 1) in the denom? i never came across a negative like that

anonymous · Answer

I don't think so... Sorry. I can't help you any further with this :( .

anonymous · Answer

that's okay, thanks for looking though

anonymous · Answer

i know igbasallote got dis

anonymous · Answer

level 99 math wizard

anonymous · Answer

I would love to see how to do this question.

raden · Answer

I GOT IT, BUT VERY LONG TO WRITTING THIS STEPS :(

raden · Answer

as information (if im not mistake), i got the answer :
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