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I was trying to prove the identity: $$\int_{0}^{\pi} {xf(\sin(x))} dx = \frac{\pi}{2} \int_{0}^{\pi} {f(\sin(x))}$$ Firstly, I substitute $x=\pi - t$. Then: $dx=-dt$ and substitute that into the integral, I get: $$I=-\int_{\pi}^{0} \pi f(\sin(\pi - t)) - tf(\sin(\pi-t))dt.$$ After that I get: $$I +\int_{0}^{\pi} tf(\sin(t)) dt=\int_{0}^{\pi} \pi f(\sin(t)) dt $$ So, can I factorize the LHS of the equation to let it become 2I, and I can prove the expression, if it can be factorized, why? The variables are different.

Thank you very much for your reply

NDewolf
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Henry Cai
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1 Answers1

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By shifting the argument,

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

becomes

$$\int_{-\pi/2}^{\pi/2} {\left(x+\frac\pi2\right)f(\cos(x))} dx = \frac{\pi}{2} \int_{-\pi/2}^{\pi/2} {f(\cos(x))}dx$$

which is true because $xf(\cos(x))$ is an odd function.

  • So, the reason why it is true is that $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} {xf(cos(x))} dx =0 $ due to the fact that the limits are the same. – Henry Cai Jan 20 '20 at 11:53
  • @HenryCai: you probably mean opposite. –  Jan 20 '20 at 11:54
  • I changed my comment, is this right? Due to the fact that cos(x) is an odd function. So, the limits are the same. – Henry Cai Jan 20 '20 at 11:55
  • @HenryCai: $\cos x$ is an even function. –  Jan 20 '20 at 12:54
  • Sorry, due to the fact that cos(x) is an even function, so cos(-x)=cos(x) so, the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} xf(cos(x)) dx$ would be equal to 0. However, there is x, x is not an even function, so, it does not equal to 0...I am quite confused, would you mind to explain this part to me? – Henry Cai Jan 20 '20 at 14:02
  • @HenryCai: it is precisely because $x$ is odd that the integral is zero. –  Jan 20 '20 at 14:46