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Integral t = tan(x/2) substitution 
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#1
Apr506, 08:25 PM

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[tex] \int \frac{dx}{5  3\sin x + 4 cos x} [/tex]
I know I have to use t = tan(x/2) substitution, and after i do that and symplify I get: [tex] \int \frac{dt}{2t^2  3t + 1} [/tex] I dont know where to go from here. If anyone can see where to go, please help. Also, if there is an easier way to do this, please tell me!! Thanks. 


#2
Apr506, 08:32 PM

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PF Gold
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One of your integral techniques is specifically for integrating fractions with polynomials in the denominator...



#3
Apr506, 08:36 PM

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P: 2,685

Sry that second integral had a 1 in the denominator instead of a 2. And that six is supposed to be a 3. I guess your saying partial fractions will work. I'll try it



#4
Apr506, 09:02 PM

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P: 2,685

Integral t = tan(x/2) substitution
ok i got an answer.
[tex]  \ln \frac{\tan(x/2)  1}{2\tan(x/2)  1)^2} + C [/tex] This seems reasonable. If anyone can find a mistake, please tell me. Thanks for the help. 


#5
Apr506, 09:06 PM

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PF Gold
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Hrm, there's a typo in your answer somewhere. Anyways, can that be simplified? The integrator at wolfram gives an answer without any logs in it.



#6
Apr506, 10:24 PM

P: 167

Hmm, I think there is something wrong with your expression [tex] \int \frac{dt}{2t^2  3t + 1} [/tex]
I tried doing the substitution myself and the expression in the denominator was a "perfect square". Maybe you can show us how you did the substitution, or you can try checking your work again... 


#7
Apr506, 11:14 PM

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P: 2,949

There is another neat way to go about it. The integration becomes easy, but the trig simplification (if desired) is slightly messy. Don't make a substitution in the first step, simply observe that (3sin x + 4cos x) = 5cos(x+arctan(3/4)).
Then the integrand can be simplified to 1/5*{1/[1 + cos(x+arctan(3/4))]}, use half angle formulae to get that to a sec^2() expression, which can be integrated immediately to a tan() expression. This is a tangent of a half angle expression, and simplifying this can be slightly (not overly) messy. But honestly, even after doing the initial substitution with the half angle tangent formula, getting it all back in terms of x is going to be messy anyway. So the way I describe is just as good. 


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