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angela210793
 4 years ago
@satellite73
angela210793
 4 years ago
@satellite73

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angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1341334653707:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ick i think the idea is to write the denominator as a single trig function

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0:O i thought maybe to split and take 2 fractions O.o

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0but idk how to do tht _

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is going to be \(5\sin(x+\theta)\) but i don't see a nice form for \(\theta=\tan^{1}(\frac{4}{3})\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually it doesn't matter because \(\theta\) is a constant also i made a mistake, it is \(5\sin(x\theta)\) still no matter

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't think there is a snappier way to do this. let me think for a second

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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 \(tan(\theta)=\frac{b}{a}\)

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0hmmmm....suppose i know tht....how to use it O.o

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well then it is easy, especially if you have a table of integrals this becomes \[\frac{1}{5}\int\csc\left(x\tan^{1}(\frac{4}{3})\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0don't worry about the arctan part, that is a number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so all you need to do is look up the anti derivative of cosecant and you are done

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here is a video explanation http://www.youtube.com/watch?v=STUh3ni4l50

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0hmmmm....we've never use secant...wht is it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cosecant it is the reciprocal of sine

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0integral 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?

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0:/....ok.. thank you Sir :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe we can get some help i will repost there might be a snappy trick

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe myininaya has a snappy way multiply top and bottom by \(3\sin(x)+4\cos(x)\) maybe?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1I don't know @satellite73 I like what you did.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok then i will stick to that. thanks

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If we replace sinx by 2tan(x/2)/(1 + tan^2(x/2),

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1yes @angela210793 sec(x)=1/(cos(x)) csc(x)=1/(sin(x))

angela210793
 4 years ago
Best ResponseYou've already chosen the best response.0ok....thanks guys :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\huge sinx = \frac{2\tan \frac{x}{2}}{1+ \tan^2\frac{x}{2}}\] \[\huge cosx = \frac{1  \tan^2\frac{x}{2}}{1+ \tan^2\frac{x}{2}}\]
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