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I have to show the following equation:

$\int_{0}^{\pi} t \cdot f(sin \; t) \; dt = \frac{\pi}{2} \int_{0}^{\pi} f(sin \; t) \; dt$

with $f : [0, 1] \rightarrow \mathbb{R}$ is continuous.

I transformed both sites to the following:

$\int_{0}^{\pi} t \cdot f(sin \; t) \; dt = \pi \cdot F(sin \; \pi)$ and

$\frac{\pi}{2} \int_{0}^{\pi} f(sin \; t) \; dt = \frac{\pi}{2} F(sin \; \pi) - \frac{\pi}{2} F(0)$

Now I don't know how to go on, because I do know nothing about $f(x)$ or $F(x)$. Is my approach correct or is there a other possibility to show the equation?

mgluesenkamp
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  • 4

1 Answers1

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HINT: $$\int_{0}^{\pi} t \cdot f(sin \; t) \; dt = \int_0^\pi (\pi-t)\cdot f(\sin(\pi-t))dt=\int_0^\pi(\pi-t)f(\sin t)dt$$

najayaz
  • 5,600