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Use the substitution in Exercise 59 to transform the integrand into a rational function of t and then evaluate the integral.

$ \displaystyle \int \frac{dx}{1 - \cos x} $

$-\cot \frac{x}{2}+C$

0:00

J H.

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

Missouri State University

Campbell University

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Okay, let's go ahead and solve this integral right here. It's actually one of the very interesting cases off into goals. Um, we do this thing called a where Stress substitution. He's a very famous mathematician, and when you talk about one of the really rigorous field off mathematics called analysis, you'll be talking about him a lot. And he's very famous. Anyhow, He figured out that no matter which kind of tribunal metric expression you have when you use this, T substitution with tangent of X over to the expression will always end up looking like a rational function and rational functions. We know a lot of techniques to solved the integral, so hopefully we can make the problem look a little bit easier is basically the idea of this thing now from the previous problems, you must have already figured out or proved what cosine or the X looks like with this substitution, you know, co sign of X equals to one minus t squared over one plus the's word, while the X is to over one plus t squared DT. So according to this substitution, we can simplify this expression with a little bit of algebra that This is the integration off one plus T squared over two teeth. This is the 1/1 minus cosine X portion, and the D X is, of course, to over one plus t's word. And as you can see, the twos cancel each other out. The one plus T's court cancels each other out, so this is simply equal to, um, one over t squared the integration off that is negative on over t, of course. And then we're going to substitute back in what T was equal to. So what's the reciprocal of tangent? It's of course, co tangent. So you will have negative co tangent off X over to Well, let's not forget the fact that this is a an indefinite, integral, so constant should always come there. And this is how you solve this integral using wire stress institution.

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