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I'm working on integration by substitution and can't seem to get a hang on the following detail:

How would one use the substitution $\displaystyle u = \pi - x$ to show the following equality:

$\int_{0}^{\pi }{x\cdot \sin \left( x \right)dx}=\; \frac{\pi }{2}\int_{0}^{\pi }{\sin \left( x \right)dx}$

My approach so far with substitution has been to find a part of the integrand to substitute for "$\displaystyle u$" and then differentiate it to get a substitution of $\displaystyle du$ for the integral $\displaystyle dx$ (and then integrate in terms of $u$); however, in using the above substitution, I cannot see how to approach this to isolate "$\displaystyle x$" from "$\displaystyle \sin(x)$" in terms of $\displaystyle u$ and $\displaystyle du$...maybe I'm missing something obvious, but can't see it.

Thanks a bunch if anyone has any insight.

Math-fun
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Topher
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  • do you know how to make $u$ substitution? $u = \pi - x$ and to convert $\int_0^\pi x \sin x , dx $ to a $u$ integral. – abel Feb 08 '15 at 12:53
  • A generalisation on this - https://math.stackexchange.com/questions/1092868/how-prove-that-int-0-pi-x-f-sin-x-dx-frac-pi-2-int-0-pi-f-si –  Dec 31 '18 at 08:02

3 Answers3

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$$I=\int_0^{\pi}x\sin x{\rm d}x\stackrel{x\to\pi-x}=\int_{\pi-0}^{\pi-\pi}(\pi-x)\sin(\pi-x){\rm d}(\pi-x)=-\int_{\pi}^{0}(\pi-x)\sin x{\rm d}x=\int_0^{\pi}(\pi-x)\sin x{\rm d}x$$ Using: $$\sin(\pi-x)=\sin x;\quad \int_a^b=-\int_b^a$$ Adding both first and last form: $$I+I=2I=\int_0^{\pi}x\sin x{\rm d}x+\int_0^{\pi}(\pi-x)\sin x{\rm d}x=\pi\int_0^{\pi}\sin x{\rm d}x\\\implies I=\frac\pi2\int_0^{\pi}\sin x{\rm d}x$$

RE60K
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  • I'm not following the use of $u = \pi - x$ here. I see you just replaced the $x$ before $sin(x)$ with the substitution, but why wouldn't you do so with the "$x$" in the sine function as well? In addition, by merely replacing x with the substitution as you did, I'm not seeing where the logic applies where this results in 2I... – Topher Feb 07 '15 at 19:16
  • @Topher $\sin(\pi-x)=\sin x$ – RE60K Feb 07 '15 at 19:17
  • $d(\pi-x)=-dx$ ! – Emilio Novati Feb 07 '15 at 19:20
  • @EmilioNovati so? – RE60K Feb 07 '15 at 19:21
  • Sorry! :-) I've confused $x$ and $u$.Upvote! – Emilio Novati Feb 07 '15 at 19:37
  • Sorry but I'm still not seeing it. Is there some integration rule/method that separates the two components $(\pi-x)$ and $sin(x)$ here, so the $(\pi-x)$ component can be solved for and moved out of the integral as a value of 1/2? Once substituted, multiplying through will give the integral of $\pisin(x)-xsin(x)$, but this results in $\pi$ times the integral of $sin(x)$ minus the integral of the original equation ($x*sin(x)$). In essence, the step done at the first arrow does not make sense to me. – Topher Feb 07 '15 at 20:08
  • @Topher, see edit. – RE60K Feb 08 '15 at 12:45
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Substituting $$x:=\pi-u,\quad dx=-du \qquad(\pi\geq u\geq0)$$ gives $$\int_0^\pi x\sin x\>dx=-\int_\pi^0(\pi -u)\sin u\>du=\pi\int_0^\pi \sin u\>du-\int_0^\pi u\sin u\>du\ ,$$ which implies $$2\int_0^\pi x\sin x\>dx=\pi\int_0^\pi \sin u\>du\ .$$

Christian Blatter
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You can integrate by parts:

$$\int_{0}^{\pi} x \sin (x)dx= [-\cos(x)x]_{0}^{\pi} + \int_{0}^{\pi} \cos(x) dx = \pi + \int_{0}^{\pi} \cos(x) dx = \pi$$.

And when we evaluate the right side, we have:

$$ \dfrac{\pi}{2} \int_{0}^{\pi} \sin(x) dx = \dfrac{\pi}{2} [-\cos(x)]_{0}^{\pi} = \pi$$

fox_lemma
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