Integral t = tan(x/2) substitution

In summary, the integral of a fraction with a polynomial in the denominator can be simplified using a substitution and half angle formulae, but the trig simplification is messy.
  • #1
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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 don't 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.
 
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  • #2
One of your integral techniques is specifically for integrating fractions with polynomials in the denominator...
 
  • #3
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
 
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  • #4
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
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
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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
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.
 

1. What is the "Integral t = tan(x/2) substitution" method used for?

The "Integral t = tan(x/2) substitution" method is used to solve integrals involving trigonometric functions, specifically those of the form ∫f(tan(x))dx. It is a technique that helps to simplify the integral and make it easier to solve.

2. How does the "Integral t = tan(x/2) substitution" method work?

The method involves substituting t = tan(x/2) into the integral, which transforms the integral into a new form that is easier to solve. This is done by using the trigonometric identity tan(x) = 2t / (1 - t^2), which is derived from the half-angle formula for tangent.

3. When should I use the "Integral t = tan(x/2) substitution" method?

This method is most useful when the integral involves trigonometric functions and cannot be solved by other methods, such as integration by parts or substitution. It is also helpful when the integral includes a secant or cosecant function, as these can be rewritten in terms of tangent.

4. Are there any restrictions when using the "Integral t = tan(x/2) substitution" method?

Yes, there are a few restrictions to keep in mind when using this method. The integral must be of the form ∫f(tan(x))dx, and the limits of integration must also be in terms of tangent. Additionally, the integral must not contain any other variable besides x and the trigonometric function.

5. Can the "Integral t = tan(x/2) substitution" method be used for all trigonometric functions?

No, this method is only applicable to integrals involving tangent, secant, and cosecant functions. For integrals involving sine and cosine, other trigonometric substitution methods like u = sin(x) or u = cos(x) should be used.

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