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I have a question regarding the following problem:

By making the substitution $x=\pi-t$, show that:

$$ \int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx $$

This is my work so far:

$$ x=\pi-t \\dx=-dt $$

$$ \int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}(t-\pi)f(\sin(\pi-t))dt\\ \int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}tf(\sin(t))dt-\pi\int_{0}^{\pi}f(\sin(t))dt $$

I noticed that $$\int_{0}^{\pi}xf(\sin{x})dx$$ and $$\int_{0}^{\pi}tf(\sin(t))dt$$ are very similar and $$\pi\int_{0}^{\pi}f(\sin(t))dt$$ is also close to what I want to find.

However, I am unsure how to make further progress.

Any help would be much appreciated

EDIT: As the comment mentioned, I forgot to change the limits of integration. After changing the limits, I got: $$ \int_{0}^{\pi}xf(\sin{x})dx=\int_{\pi}^{0}tf(\sin(t))dt-\pi\int_{\pi}^{0}f(\sin(t))dt\\ \int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\pi}f(\sin(t))dt-\int_{0}^{\pi}tf(\sin(t))dt $$

I can see that this is very close to the final answer. However, I still don't know how I can eliminate the variable $t$

FarmerZee
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  • Careful - when you make the change of variables your limits of integration should change. You're very close. – DMcMor Jan 15 '21 at 18:42

4 Answers4

4

Just a few sign flips away from it. $$\int_{0}^{\pi}xf(\sin{x})dx=\int_{\pi}^{0}(\pi-t)f(\sin(\pi-t))(-dt)\\ =\int_{0}^{\pi}(\pi-t)f(\sin(\pi-t))dt\\ =\pi\int_{0}^{\pi}f(\sin(t))dt-\int_{0}^{\pi}tf(\sin(t))dt $$

Clearly, the value of a definite integral is independent of the dummy variable we use to describe it. So we can do an algebraic manipulation to get

$$2\int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\pi}f(\sin(t))dt$$ $$\int_{0}^{\pi}xf(\sin{x})dx=\frac\pi2\int_{0}^{\pi}f(\sin(t))dt$$

2

You have made a mistake - your last equation is missing a negative sign. If you let $I=\displaystyle \int_0^{\pi} x f(\sin x)\mathrm dx$ , then the correct result is $$\int_0^{\pi} xf(\sin x)\text dx=\pi \int_0^{\pi}f(\sin t) \text d t-\int_0^{\pi} tf(\sin t)\text d t$$ But since $$\displaystyle \int_0^{\pi} xf(\sin x)\text dx=\int_0^{\pi }tf(\sin t) \text d t$$ because $t$ is just a dummy variable, we get $\displaystyle I=\dfrac{\color{red}\pi}{2} \int_0^{\pi} f(\sin x) \text{d} x$.

Edit: This was answered in context to original question before editing.

Teymour
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V.G
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  • I don't quite understand the concept of the dummy variable. If we previously defined x=pi-t, how can let t=x again? – FarmerZee Jan 15 '21 at 19:07
  • because it doesn't matter what variable we use for integration, we always have $\displaystyle \int_{a}^b f(x) \text dx=\int_a^b f(u) \text d u$. – V.G Jan 15 '21 at 19:08
  • @FarmerZee: Read this answer. – V.G Jan 15 '21 at 19:10
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hint

Let $$I=\int_0^\pi xf(\sin(x))dx$$

with $ t=\color{red}{\pi} -x$, it becomes

$$I=\int_{\color{red}{\pi}-0}^{\color{red}{\pi}-\pi}(\pi-t)f(\sin(\pi-t))(-dt)=$$

$$\int_0^\pi(\pi-t)f(\sin(t))(+dt)=$$

$$\pi\int_0^\pi f(\sin(t))dt-I$$

thus

$$2I=\pi \int_0^\pi f(\sin(t))dt$$ $$=\pi \int_0^\pi f(\sin(x))dx$$

and $$I=\frac{\pi}{2}\int_0^\pi f(\sin(x))dx$$

hamam_Abdallah
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$$I=\int_0^\pi x\,f(\sin x)dx$$ $x=\pi-t\Rightarrow dx=-dt$ so: $$I=\int_\pi^0(\pi-t)f\left[\sin(\pi-t)\right](-dt)=\int_0^\pi(\pi-t)f(\sin t)dt$$ $$=\pi\int_0^\pi f(\sin t)dt-\int_0^\pi t\,f(\sin t)dt$$ which is the same thing as saying: $$I=\pi\int_0^\pi f(\sin t)dt-I$$ and so rearrange this equation and we get: $$I=\frac\pi2\int_0^\pi f(\sin x)dx$$

Henry Lee
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