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It is also given that $f$ is a continuous function.

Specifically, the question tasks to use the substitution $x=\pi - t$. However, I don't seem to be able to get anything useful out of this substitution:

$$ \begin{align} \int_0^\pi xf(\sin x)dx&=\int_\pi^0 (\pi -t)f(\sin (\pi - t)) (-dt)\\&=-\int_\pi^0 (\pi -t)f(\sin (t))dt\\&=\int_0^\pi (\pi -t)f(\sin (t))dt \end{align} $$

Then:

$$\begin{align}\int_0^\pi (\pi -t)f(\sin (t))dt&=\int_0^\pi \pi f(\sin (t))dt - \int_0^\pi tf(\sin (t))dt\\&=\pi\int_0^\pi f(\sin (t))dt - \int_0^\pi tf(\sin (t))dt \end{align} $$

However here is where I'm lost. By putting back $t=\pi-x$ I'm just reversing the process which puts me right back where I started, leaving me with no clue on how to continue. I also tried to use integration by parts on the last result, but it didn't look like it would yield anything of value.

My gut is telling me that I should be able to somehow get something in the lines of:

$\int_0^\pi xf(\sin x)dx=\pi\int_0^\pi f(\sin (x))dx - \int_0^\pi xf(\sin (x))dx$

then I would move $\int_0^\pi xf(\sin (x))dx$ to the other side and get the desired result. But I'm stuck with the substituted $t$ variable instead.

Any hints would be appreciated, thanks in advance.

NOTE: I've seen other answers which use:

$\int_0^\pi f(x)dx=\int_0^\pi f(0+\pi-x)dx$

but the question specifically asks to use $x=\pi - t$.

Thomas Andrews
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Emanuel L
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  • https://math.stackexchange.com/q/639443/42969, https://math.stackexchange.com/q/971888/42969, https://math.stackexchange.com/q/3986712/42969 – Martin R Apr 07 '24 at 18:26
  • Thanks Martin, but as I've said I've already peeked at some of the answers and saw them using the formula I specified at the bottom of the post. The question specifically asks to use $x=\pi - t$. Maybe I'm missing the connection between the two. – Emanuel L Apr 07 '24 at 18:29
  • Your hunch is absolutely right in how you derive the identity. Just because the variable of integration is labelled as $t$ instead of $x$ doesn't mean that the integrals are not equal. It's the same integrand over the same bounds. – chirpyboat73 Apr 07 '24 at 18:29
  • $\int_0^\pi f(x)dx=\int_0^\pi f(0+\pi-x)dx$ is the substitution $x=\pi -t $, then renaming $t$ to $x$. – Martin R Apr 07 '24 at 18:29
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    Note that $$\int_a^b f(x),dx=\int_a^b f(t),dt=\int_a^b f(u),du=\int_a^b f(\text{anything}),d,\text{anything}$$ – Mark Viola Apr 07 '24 at 18:45
  • @MarkViola thanks for the comment. But how can both of the integrals in your comment be equal when $x=\pi - t$? – Emanuel L Apr 07 '24 at 18:58
  • To @MarkViola's point, take a look at How do I convince my students that the choice of variable of integration is irrelevant? – user170231 Apr 07 '24 at 19:21
  • On one hand it is obvious to me that the equalities mentioned in @MarkViola 's comment are correct. I understand them. I don't understand how the equality $\int_0^\pi xf(\sin x)dx = \int_0^\pi tf(\sin t)dt$ is correct, if I previously substituted $x=\pi-t$... – Emanuel L Apr 07 '24 at 19:40
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    There are solutions in the links provided in the comments, although I prefer the following. The function $tf(\cos t)$ for $|t|\le \pi/2$ is odd therefore its integral over $[-\pi/2,\pi/2]$ vanishes. Substitution $t=x-\pi/2$ gives $$\int\limits_0^\pi (x-\pi/2)f(\sin x),dx =0$$ and the conclusion follows. – Ryszard Szwarc Apr 07 '24 at 20:11
  • @EmanuelL In my comment, let that function $f(x)$ be your function $xf(x)$. You see, the variable that appears under the definite integral is a "dummy" variable. It is a symbol only. One may replace any one symbol with any other symbol and the value of the integral is unaffected. – Mark Viola Apr 08 '24 at 14:38

1 Answers1

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Hint: Let $I=\int_0^\pi xf(\sin x)\,dx$. Then after the substitution $t=\pi-x$, you get $$I=\int_0^\pi xf(\sin x)\,dx=\pi\int_0^\pi f(\sin x)\,dx-\int_0^\pi xf(\sin x)\,dx=\pi\int_0^\pi f(\sin x)\,dx-I.$$ Now solve for $I$!

csch2
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  • This will probably sound dumb, but where did $t$ go to? – Emanuel L Apr 07 '24 at 18:46
  • Is that solution different from those in the three Q&As that I linked to in my first comment? – Martin R Apr 07 '24 at 18:47
  • Sorry Martin, the solution may be apparent to you, and in your eyes this answer is probably identical to all others. I did follow the links, read the answers and followed the advised links in the comments. I would elaborate about what I don't understand in those links, but I think it would be too long for a comment. – Emanuel L Apr 07 '24 at 18:56