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I am preparing for a basic level calculus test and came across this problem:

$$ \int_0^{\pi/2} \dfrac{1}{2-\cos{x}} dx$$

Which appears to be simple enough before I realize that I can't multiply the top and bottom by $ 2+\cos{x} $ and get anywhere fast. The prompt and Wolfram Alpha suggest a u-substitution of $ u = \tan{\frac{x}{2}} $ but I really have no idea how to implement that. How is this substitution used? Is this even an appropriate problem for my current level of math knowledge?

intcreator
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    The $u=\tan \frac x2$ substitution is also called the Weirstrass substitution; it's widely-used because all of the trigonometric functions of $x$ are rational functions of $\tan \frac x2$. See http://en.wikipedia.org/wiki/Tangent_half-angle_substitution for more details. – Steven Stadnicki Mar 03 '15 at 05:47
  • Ok I'm pretty sure we have definitely not covered that so far in my class so I'm going to stop worrying about not having known what it was. Thanks for the link - that really answered my question. – intcreator Mar 03 '15 at 05:53

2 Answers2

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hint: use the sub $u = \tan( x/2), \cos t = \frac{1-u^2}{1+u^2}$ and some partial fraction.

differencing you get $$du = (1 + u^2) dx/2, x = 0 \to u = 0, x = \pi/2\to u = 1$$ therefore the $\int_0^{\pi/2} \frac{dx}{2-\cos x}$ is transformed into $$\int_0^1 \frac{2du}{(1 + u^2} \frac{1}{\left(2 - \frac{1-u^2}{1+u^2}\right)} =2 \int_0^1\frac{du}{3u^2 + 1} = \frac{2}{3} \int_0^1 \frac{du}{u^2 + 1/3}$$ can you continue now?

abel
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  • Okay but how am I supposed to get cos t = (1-u^2)/(1+u^2) from u = tan(t/2) ? That's where I'm completely lost. – intcreator Mar 03 '15 at 05:32
  • @Brandon, i will edit my post. show some of the steps. – abel Mar 03 '15 at 05:34
  • I'm trying to follow along but I'm still not sure how the cos x gets transformed like that. Perhaps I'm simply not familiar that substitution. Other than that the rest of the equation makes sense. How am I supposed to know what to put in the place of cos x? Is it simply a well-known substitution? – intcreator Mar 03 '15 at 05:49
  • @Brandon, here is way to see that. you need the half/double ante formula: $\cos x = \cos^2 x/2 - \sin x/2 = \frac{\cos^2 x/2 - \sin^2 x/2}{\cos^2 x/2 + \sin^2 x/2}$ now divide both numerator and denominator by $\cos^2 x/2.$ it gives you $\cos x = \frac{1 - \tan^2 x/2}{1 + \tan^2 x/2}$ – abel Mar 03 '15 at 05:53
  • Oh okay cool. That makes a lot more sense. I've actually never seen that formula before. – intcreator Mar 03 '15 at 05:57
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We start with $$\int_0^{\pi/2} \dfrac{1}{2-\cos{x}} dx$$ Let's start as you suggest and see where it takes us, multiplying top and bottom by $2 + \cos x$. $$ \int_0^{\pi/2} \frac{2 + \cos x}{4 - \cos^2 x} dx = \int_0^{\pi/2} \frac{2 + \cos x}{3 + \sin^2 x} dx.$$ The mathematics of wishful thinking suggests that we split this integral and handle the easy part first. $$ \int_0^{\pi/2} \frac{\cos x}{3 + \sin^2 x} dx = \int_0^1 \frac{1}{3 + u^2} du$$ when we perform the substitution $u = \sin x$. This is now a classic inverse trigonometric integral, and will lead you to $\arctan$.

What did we leave behind? We left $$ \int_0^{\pi/2} \frac{2}{3 + \sin^2 x} dx.$$ This looks a bit tricky. One way to proceed is to recognize $3 = 3\sin^2 x + 3 \cos^2 x$, so that we have $$ \int_0^{\pi/2} \frac{2}{3 \cos^2 x + 4\sin^2 x} dx.$$ Now multiplying by $\sec^2(x)/\sec^2(x)$ gives $$\int_0^{\pi/2} \frac{2 \sec^2 x}{3 \cot^2 x + 4} dx.$$ Now you can finish with the substition $u = 3\cot^2 x + 4$.

Is this what I'd recommend? It's hard to say. The substitution you mention is a well-known substitution. But it's important to recognize that with these problems, as long as you follow your nose and charge forward, you'll probably reach a resolution.

davidlowryduda
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  • $(2-\cos x)(2+\cos x)=4-\cos^2 x$, not $4+\cos^2 x$ - but then that does equal $3+\sin^2 x$ so the rest goes through... – Steven Stadnicki Mar 03 '15 at 05:42
  • @StevenStadnicki oh, well what a silly mistake to start with – davidlowryduda Mar 03 '15 at 05:43
  • Apologies for the tone of my first post - I hadn't realized it was just a sign error there rather than a deeper error. This is a nice approach all in all (+1), although I'd probably never find the final transformation myself. – Steven Stadnicki Mar 03 '15 at 05:49
  • Thanks for the alternate approach. I just didn't go through with it enough when I tried it (and made some errors as well which didn't help). – intcreator Mar 03 '15 at 05:58