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ick i think the idea is to write the denominator as a single trig function
:O i thought maybe to split and take 2 fractions O.o

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Other answers:

but idk how to do tht -_-
it is going to be \(5\sin(x+\theta)\) but i don't see a nice form for \(\theta=\tan^{-1}(\frac{4}{3})\)
actually it doesn't matter because \(\theta\) is a constant also i made a mistake, it is \(5\sin(x-\theta)\) still no matter
i don't think there is a snappier way to do this. let me think for a second
okok
ok no i can't think of a better way, and also i tried wolfram and what a disaster what you need to know is \[a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\] where \(tan(\theta)=\frac{b}{a}\)
hmmmm....suppose i know tht....how to use it O.o
well then it is easy, especially if you have a table of integrals this becomes \[\frac{1}{5}\int\csc\left(x-\tan^{-1}(\frac{4}{3})\right)\]
don't worry about the arctan part, that is a number
so all you need to do is look up the anti derivative of cosecant and you are done
here is a video explanation http://www.youtube.com/watch?v=STUh3ni4l50
hmmmm....we've never use secant...wht is it?
cosecant it is the reciprocal of sine
integral is \[\int csc(x)dx=-\ln(\cot(x)+\csc(x))\] if you have not used this i really have no idea how you are supposed to do this problem maybe multiply top and bottom by the conjugate?
:/....ok.. thank you Sir :D
maybe we can get some help i will repost there might be a snappy trick
okk
maybe myininaya has a snappy way multiply top and bottom by \(3\sin(x)+4\cos(x)\) maybe?
I don't know @satellite73 I like what you did.
ok then i will stick to that. thanks
I actually think what sat did was probably the most snappiest thing you can do for this type of integral
If we replace sinx by 2tan(x/2)/(1 + tan^2(x/2),
sec=1/cosx????
yes @angela210793 sec(x)=1/(cos(x)) csc(x)=1/(sin(x))
ok....thanks guys :D
\[\huge sinx = \frac{2\tan \frac{x}{2}}{1+ \tan^2\frac{x}{2}}\] \[\huge cosx = \frac{1 - \tan^2\frac{x}{2}}{1+ \tan^2\frac{x}{2}}\]

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