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I solved this integral: $$ \int_0^{\pi/2}\frac{dx}{5+4\cos x} $$ and I obtained $\frac{2}{3} \arctan \frac{1}{3}$ is it correct? how do I find the answer in $\pi$?

Michael Hardy
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Lola
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  • use the Weierstrass substitution – Dr. Sonnhard Graubner May 19 '17 at 15:39
  • I did and I found that answer?how do I find the answer in pi? – Lola May 19 '17 at 15:40
  • $$\int_{0}^{4\pi}\frac{dt}{5+4\cos t}=\frac{4\pi}{3}$$ but there is no way to guess the correct integration bounds from your question. – Jack D'Aurizio May 20 '17 at 13:04
  • @Dr.SonnhardGraubner : "Weierstrass" would appear to be a misnomer. James Stewart's calculus textbooks assert that Weierstrass introduced that substitution, and Stewart said elsewhere than in his books that the name was around before he asserted that. I suspect Stewart uncritically assumed that if people called it that then it's a historical fact that Weierstrass introduced it. History was not Stewart's field of scholarly interest. http://en.wikipedia.org/wiki/Tangent_half-angle_substitution – Michael Hardy May 21 '17 at 17:18
  • $$\begin{align} & \int_0^{\pi/2} \frac{dx}{5+4\cos x} = \int_0^1 \frac{\left( \frac {2,dt}{1+t^2} \right) }{5 + 4 \left( \frac{1-t^2}{1+t^2} \right)} \[15pt] = {} & \int_0^1 \frac{2,dt}{5(1+t^2) + 4(1-t^2)} = \int_0^1 \frac{2,dt}{9+t^2} \[15pt] = {} & \left. \frac 2 3 \arctan \frac t 3 \right|_0^1 = \frac 2 3 \arctan \frac 1 3 \end{align}$$ – Michael Hardy May 21 '17 at 17:36
  • Do you mean express $\frac23\arctan\frac13$ as some simple multiple of $\pi$? This is impossible algebraically. In general, we know for any rational $x \ne 0, \pm 1$, $\frac{1}{\pi}\arctan(x)$ is a transendental number. This is first pointed out by Margolius on hir study of transcendence of Plouffe's constant. – achille hui May 21 '17 at 18:28

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and I obtained $\frac{2}{3} \arctan \frac{1}{3}$ is it correct?

Yes, that is correct!

How do I find the answer in $\pi$?

What do you mean, "in $\pi$"?

Perhaps the model answer looks different (containing $\pi$)? You can use the property: $$\boxed{\arctan a + \arctan \frac{1}{a} = \frac{\pi}{2}} \implies \color{blue}{\arctan\frac{1}{3} = \frac{\pi}{2}-\arctan 3 }$$ to rewrite as follows: $$\frac{2}{3}\color{blue}{\arctan\frac{1}{3}} = \frac{2}{3}\left(\color{blue}{ \frac{\pi}{2}-\arctan 3} \right) = \frac{\pi}{3}-\frac{2}{3}\arctan 3$$

StackTD
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  • I think that the OP means the value of $arctan \frac{1}{3}$ expressed in terms of $\pi$, which, is not possible to obtain. – Ananth Kamath May 19 '17 at 15:47
  • I tried using that property but the problem is that the answers in my textbooks dont contain arctan at all :( ,the answers are: $0$ $\frac{4pi}{3}$ $\frac{4}{5}pi$ ,$pi$, $\frac{5}{4}pi$ – Lola May 19 '17 at 15:48
  • I think they got the answers wrong – Lola May 19 '17 at 15:49
  • @Lola You can rewrite it in different ways, but it's not equal to any of the answers you (or they...) suggested. Perhaps a mistake in the answers, or they meant a different integral. – StackTD May 19 '17 at 15:50
  • I just read on a forum that the integral was meant to be from 0 to 4pi in the textbook.But what do I do if I have 4pi,I know its a lot like the others integrals on the site but I just don't understand how to use Weierstrass in this conditions – Lola May 19 '17 at 15:56
  • I would guess that "in $\pi$" means "as a function of $\pi$". $\qquad$ – Michael Hardy May 21 '17 at 17:19