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I'm helping my daughter on her calculus homework and it has been many years for me. The problem is $$ \int \frac{1}{2+3 \sin\left(x\right)} dx $$ From WolframAlpha, the substitution should be $u = \tan\frac{x}{2}$. Once you have the substitution, the solution is quite straightforward.

My question is how on earth do you come up with that substitution? Are there any rubrics/recipes? Or is it just intuition?

Tunococ
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user90855
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    The Weierstrass substitution is a common trick to evaluate trigonometric integrals. It is not something you (or your daughter) are supposed to some up with yourself; you are supposed to recognise the trigonometric form, know that such a substition might work, and try to see if it does. That said, there may of course be other (and better/easier) solutions in some cases. – Myself Jul 12 '13 at 20:52
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    Adding to what Myself just wrote, the Wikipedia article notes that " Michael Spivak wrote that "The world's sneakiest substitution is undoubtedly" this technique" – Baby Dragon Jul 12 '13 at 20:54

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Weierstrass Substitution, $z=\tan(x/2)$, works well in this kind of integral: $$ \begin{align} \sin(x)&=\frac{2z}{1+z^2}\\[6pt] \cos(x)&=\frac{1-z^2}{1+z^2}\\[6pt] \mathrm{d}x&=\frac{2\mathrm{d}z}{1+z^2} \end{align} $$ Then $$ \begin{align} \int\frac{\mathrm{d}x}{2+3\sin(x)} &=\int\frac1{2+3\frac{2z}{1+z^2}}\frac{2\mathrm{d}z}{1+z^2}\\ &=\int\frac{\mathrm{d}z}{z^2+3z+1}\\ &=\int\frac{\mathrm{d}z}{(z+3/2)^2-5/4}\\ &=\frac1{\sqrt5}\int\left(\frac1{z+3/2-\sqrt{5/4}}-\frac1{z+3/2+\sqrt{5/4}}\right)\,\mathrm{d}z \end{align} $$ which is a simple form (thanks to Michael Hardy for pointing out partial fractions). Then back-substitute to get the answer in terms of $x$.

robjohn
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  • Instead of that last trigonometric substitution, you could just use partial fractions after you've got an integral of a rational function. – Michael Hardy Jul 12 '13 at 22:18
  • Doh! I will fix that. Thanks! – robjohn Jul 12 '13 at 22:26
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    It's also been questioned on mathoverflow recently whether the "Weierstrass solution" should be called that. Fred Rickey has said he's searched through Weierstrass' publications without finding it. Stewart's Calculus calls it that, IIRC. So did Stewart introduce the usage, or is it older? – Michael Hardy Jul 12 '13 at 22:54
  • @MichaelHardy: I don't know the history. I just go by what Wikipedia calls it. I'd be interested to know if it has a better name since I use it a lot. – robjohn Jul 12 '13 at 23:00
  • I can't go by what Wikipedia calls it, since I am the initial creator of that Wikipedia article (and hundreds of others), and I just went by what some book called it. I think it might have been Stewart. – Michael Hardy Jul 12 '13 at 23:01
  • @MichaelHardy: before I saw the article on Wikipedia, I used to call it the "half tangent" or "tangent half angle" substitution. I see that on Wikibooks, they call it the tangent half angle substitution, even in the September 2008 revision. – robjohn Jul 12 '13 at 23:06
  • The origin of the substitution is the rational parametrization of the unit circle (which surely must predate Weierstrass). Consider the line with slope $u$ through $(-1,0)$. Its other intersection with the unit circle is at $(\cos\theta,\sin\theta)$, and so $u=\tan(\theta/2)$. – Ted Shifrin Jul 13 '13 at 01:33
  • Correction: It was here on m.s.e. that this was questioned. – Michael Hardy Jul 13 '13 at 15:12
  • @MichaelHardy: I hope you don't mind me adding the link to your comment. – robjohn Jul 13 '13 at 15:54
  • It was one of my posted questions: http://math.stackexchange.com/questions/420364/who-is-buried-in-weierstrass-tomb – Michael Hardy Jul 13 '13 at 17:47