I am stuck on the integral.
$$\int_0^\pi \frac{x \sin x}{1+ \cos^2x} dx$$
The hint was to swap limits and add something to cancel the $x$ out? If the denominator was $1 - \cos^2x$ we'd be in business lol.
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2This integral has a horrible exact form, and evaluating it (according to Wolfram) from $0$ to $\pi$ gives $\pi^2/4$ – FShrike Aug 18 '21 at 19:06
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1If there was no $;x;$ in the numerator, as they hint, you'd be in business as you'd have an integral of the form $;\int\frac{f'}{1+f^2} dx=\arctan f;$ ...in fact, having$;1-\cos^2x=\sin^2x;$ in the denominator would render a rather horrible integral... – DonAntonio Aug 18 '21 at 19:07
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@FShrike but how do do it without Wolfram lol I'm studying for GRE subject math this came up on U of Chi practice probems – MyMathYourMath Aug 18 '21 at 19:09
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@DonAntonio so what do I do lol as there is an x. any idea on how to do this one – MyMathYourMath Aug 18 '21 at 19:10
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1Keep the following in mind: GRE problems are by and large not overly difficult. They may however involve a trick to reduce them to something much simpler. Symmetries (such as $x\mapsto \pi-x$) are a good thing to try if you think the problem is very difficult at face value. – Cameron L. Williams Aug 18 '21 at 19:53
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@CameronWilliams I was told this problem set is by far the most difficult its from uni of Chicago math department. im doing week 3 atm – MyMathYourMath Aug 18 '21 at 19:54
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1Does this answer your question? Evaluate $\int^{\pi}_0\frac{x\sin(x)}{1+\cos^2(x)}dx$ – Axion004 Aug 18 '21 at 20:15
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Also a duplicate of this question, this question, this question,and this question. – Axion004 Aug 18 '21 at 20:17
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And more duplicates: this question, and this question. – Axion004 Aug 18 '21 at 20:18
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I have added an answer here. I haven't looked through all the answers to all the duplicates to see if this is the same as another answer, but it is different than the ones I've seen so far. – robjohn Aug 19 '21 at 11:03
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The answer I added is almost the same as this answer. – robjohn Aug 19 '21 at 11:08
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@robjohn: This answer is identical to the accepted answer here, it is also the same as the answer provided by Harish here, and the answer provided by mattos here. – Axion004 Aug 19 '21 at 14:32
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The same technique is also shown here and in both of the answers here. – Axion004 Aug 19 '21 at 14:34
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@Axion004: are you talking about the answer to this question or the answer I added? I found only one answer that was close to the answer I added. – robjohn Aug 19 '21 at 14:35
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Your answer is different than all of the others. The accepted answer below is the same as the approach in the duplicate questions. – Axion004 Aug 19 '21 at 14:37
1 Answers
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Denote by $$I := \int_0^\pi \frac{x \sin(x)}{1+\cos(x)^2}dx. $$
With the change of variables $x = \pi - t$, we get that $$I = \int_0^\pi \frac{(\pi - t) \sin(\pi - t)}{1 + \cos(\pi-t)^2}dt. $$
As $\sin(\pi-t) = \sin(t)$ and as $\cos(\pi - t) = -\cos(t)$, we have that $$I = \int_0^\pi \frac{(\pi - t) \sin(t)}{1 + \cos(t)^2}dt = \pi \int_0^\pi \frac{\sin(t)}{1 + \cos(t)^2}dt - I, $$ and so $$I = \frac{\pi}{2} \int_0^\pi \frac{\sin(t)}{1 + \cos(t)^2}dt. $$
The last integral can be computed using the change of variable $\cos(t) = y$, as mentioned in the comments.
C_M
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7It is true that $dx = -dt$, but the change of variables $x = \pi - t$ flips the limits of integration, and these two cancel out. – C_M Aug 18 '21 at 19:26
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1Aha! Then this is a good answer with good technique. I wonder why it can't be generalised and Wolfram's monster is necessary for a general integral – FShrike Aug 18 '21 at 19:29
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so after evaluating the integral ill get the desired result? @C_M – MyMathYourMath Aug 18 '21 at 19:32
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1Yes, the final answer is $\pi^2/4$, as @FShrike mentioned in the comments. – C_M Aug 18 '21 at 19:34
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1This is an excellent, helpful answer. The last integral is almost immediate and wouldn't even require change of variables: the integrand is $;\cfrac{f'}{1+f^2};$ , so the integral equals $;\arctan f;$ .... (chain rule backwards) – DonAntonio Aug 18 '21 at 19:59
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Such integrals are quite easy, imo, if you have enough experience.(+1) – Laxmi Narayan Bhandari Aug 19 '21 at 02:51
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@FShrike : (about Wolfram's complicated answer) This integral contains a special symmetry owing to the particular bounds from $0$ to $\pi$. The answer Wolfram gives is for generic limits, which may not be symmetric. – The_Sympathizer Aug 19 '21 at 06:08
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This is a reminder to refrain from answering low-quality questions. See Enforcement of Quality Standards. – TheSimpliFire Aug 19 '21 at 06:37