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Show that $\int \limits_{0}^{\pi} f(\sin x) \cos x \, dx =0$ for any function $f$ continuous on $[0,1]$.

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    $\begingroup$ Hint: change of variable $u=\sin x$. $\endgroup$ – Did Mar 28 '14 at 23:18
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    $\begingroup$ Is this intuitively true to anyone? $\endgroup$ – Git Gud Mar 28 '14 at 23:21
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    $\begingroup$ This would be a nice calculus exam question, with some disguise e.g. $\int_0^{\pi} (\sin x)^{\log(1+\sin x)}\cos x\mathrm{d}x$ $\endgroup$ – Lord Soth Mar 28 '14 at 23:28
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    $\begingroup$ @GEdgar: Fortunately, there is no need to invert the substitution! $\endgroup$ – user14972 Mar 29 '14 at 0:11
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    $\begingroup$ @GitGud After a few seconds thought, yes, sort of. Each value that is taken by $\sin(x)$ is taken twice, but with negative values for $\cos(x)$. So the function takes its positive values and the countering negative values with equal measure. $\endgroup$ – alex.jordan Mar 29 '14 at 1:06
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First, we split up the integral into $\int_0^{\pi/2}{f(\sin x)\cos x dx}+\int_{\pi/2}^\pi{f(\sin x) \cos x dx}$, and then notice that by letting $\pi-y=x$, we have $$\int_{\pi/2}^\pi{f(\sin x)\cos x dx}=\int_{\pi/2}^{0}{-f(\sin (\pi-y)) \cos (\pi-y)dy}=-\int_0^{\pi/2}{f(\sin y)\cos y dy}.$$ Thus, we find that $$\int_0^\pi{f(\sin x)\cos x dx}=\int_0^{\pi/2}{f(\sin x)\cos x dx}-\int_0^{\pi/2}{f(\sin x)\cos x dx}=0$$

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    $\begingroup$ Up voted for originality, but I prefer Did's solution. $\endgroup$ – Git Gud Mar 28 '14 at 23:53
  • $\begingroup$ Figured I should make some kind of positive contribution after making dumb mistakes. $\endgroup$ – Hayden Mar 28 '14 at 23:56
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    $\begingroup$ I prefer this solution since the symmetries of $\sin$ and $\cos$ are well known and immediatly imply this result. $\endgroup$ – Christoph Mar 28 '14 at 23:56
  • $\begingroup$ I actually prefer this solution as well, the u-substitution is certainly easier, but the result seems strange (how can one integral over a non-zero domain transform to an integral from $0$ to $0$--it does, but I think it warrants a double take). $\endgroup$ – Jared Mar 29 '14 at 0:09
  • $\begingroup$ @Jared: It's no more weird than tracing out a closed curve; and the sine function on $[0, \pi]$ is a closed curve in $\mathbf{R}$. :) $\endgroup$ – Andrew D. Hwang Mar 29 '14 at 0:40
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Since $f$ is continuous on $[0,1]$, there exists $F$ on $[0,1]$ such that $f=F'$, then

$$\int_0^\pi f(\sin x)\cos x\ dx=\int_0^\pi(F(\sin x))'\ dx=[F(\sin x)]_0^\pi=F(0)-F(0)=0$$

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    $\begingroup$ Or even more briefly: By Did's substitution and change of variables, $$\int_0^\pi f(\sin x)\cos x\, dx = \int_0^0 f(u)\, du = 0.$$ $\endgroup$ – Andrew D. Hwang Mar 29 '14 at 0:41
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Substitute $y=x-\frac {\pi}2$ and you will find that you are integrating an odd function with symmetric limits, which is trivially zero.

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