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angela210793Best ResponseYou've already chosen the best response.0
dw:1341334653707:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
ick i think the idea is to write the denominator as a single trig function
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
:O i thought maybe to split and take 2 fractions O.o
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
but idk how to do tht _
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
it is going to be \(5\sin(x+\theta)\) but i don't see a nice form for \(\theta=\tan^{1}(\frac{4}{3})\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
actually it doesn't matter because \(\theta\) is a constant also i made a mistake, it is \(5\sin(x\theta)\) still no matter
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
i don't think there is a snappier way to do this. let me think for a second
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
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 \(tan(\theta)=\frac{b}{a}\)
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
hmmmm....suppose i know tht....how to use it O.o
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
well then it is easy, especially if you have a table of integrals this becomes \[\frac{1}{5}\int\csc\left(x\tan^{1}(\frac{4}{3})\right)\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
don't worry about the arctan part, that is a number
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
so all you need to do is look up the anti derivative of cosecant and you are done
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
here is a video explanation http://www.youtube.com/watch?v=STUh3ni4l50
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
hmmmm....we've never use secant...wht is it?
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
cosecant it is the reciprocal of sine
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
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?
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
:/....ok.. thank you Sir :D
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
maybe we can get some help i will repost there might be a snappy trick
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
maybe myininaya has a snappy way multiply top and bottom by \(3\sin(x)+4\cos(x)\) maybe?
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
I don't know @satellite73 I like what you did.
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
ok then i will stick to that. thanks
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
I actually think what sat did was probably the most snappiest thing you can do for this type of integral
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
If we replace sinx by 2tan(x/2)/(1 + tan^2(x/2),
 one year ago

myininayaBest ResponseYou've already chosen the best response.1
yes @angela210793 sec(x)=1/(cos(x)) csc(x)=1/(sin(x))
 one year ago

angela210793Best ResponseYou've already chosen the best response.0
ok....thanks guys :D
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
\[\huge sinx = \frac{2\tan \frac{x}{2}}{1+ \tan^2\frac{x}{2}}\] \[\huge cosx = \frac{1  \tan^2\frac{x}{2}}{1+ \tan^2\frac{x}{2}}\]
 one year ago
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