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$$I=\int\frac{1}{2+\cos\theta}d\theta$$

When trying to solve this integral my calculus-book states that we can use the following substitution:

$$x=\tan(\theta/2)$$$$\cos\theta=\frac{1-x^2}{1+x^2}$$$$d\theta=\frac{2dx}{1+x^2}$$

This substitution results in this easier to solve integral:

$$I=\int\frac{2}{3+x^2}dx$$

My question is about the validity of this substitution for values where $x$ does not exist, for example at $\theta=\pi$, at which $\tan(\theta/2)$ is not defined.

I don't understand how the substitution can be valid in general while $x$ is not defined for all values of $\theta$.

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See the Geometry part of wiki thread. It describes the parameterization in detail. In particular, it says the point never reaches the point $(−1,0)$. So you're correct. $x$ is not defined when $\theta$ approached $\pi$.

Intuitively, for me, any tri-sub(including tangent substitution) is just a kind of "transformation".

Hope this makes some sense to you.

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  • $\begingroup$ Ah, the second example on the wiki site basically answers my question. $\endgroup$ – GambitSquared Apr 21 '18 at 19:16
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first find out the final function after integration. you notice that if you put θ= pi in the function the answer is not undefined.

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  • $\begingroup$ That feels a bit unsatisfactory: so we just happen to find out it works regardless? There must be a deeper explanation of what is going on here? $\endgroup$ – GambitSquared Apr 21 '18 at 18:50
  • $\begingroup$ oh k so you want to know why all the substitution like this won`t be undefined.? $\endgroup$ – Avnish Singh Apr 21 '18 at 18:51
  • $\begingroup$ that would require some thinking, ha $\endgroup$ – Avnish Singh Apr 21 '18 at 18:52
  • $\begingroup$ "its the beauty of maths" would be a vague answer, i guess $\endgroup$ – Avnish Singh Apr 21 '18 at 18:52
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The substitution isn't valid, in general, due to not every $x$ mapping to a value of $\theta$. When it isn't valid, you should find inconsistencies. When this occurs, use a different substitution.

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