Integral t = tan(x/2) substitution

1. Apr 5, 2006

G01

$$\int \frac{dx}{5 - 3\sin x + 4 cos x}$$

I know I have to use t = tan(x/2) substitution, and after i do that and symplify I get:

$$\int \frac{dt}{2t^2 - 3t + 1}$$

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.

Last edited: Apr 5, 2006
2. Apr 5, 2006

Hurkyl

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

3. Apr 5, 2006

G01

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

Last edited: Apr 5, 2006
4. Apr 5, 2006

G01

ok i got an answer.

$$| \ln \frac{\tan(x/2) - 1}{2\tan(x/2) - 1)^2} + C$$

This seems reasonable. If anyone can find a mistake, please tell me. Thanks for the help.

5. Apr 5, 2006

Hurkyl

Staff Emeritus
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. Apr 5, 2006

pizzasky

Reply

Hmm, I think there is something wrong with your expression $$\int \frac{dt}{2t^2 - 3t + 1}$$

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. Apr 5, 2006

Curious3141

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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