# Integral equality

https://imgur.com/u0rL1EV

Is it true that

$\int_{-\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx?$
• Dr Ali
+2

It is better to type body for Pro Bono questions and do not use links or images

• I edited the question.

The answer is true. We have

$\int_{-\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx+\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx$
$=0+\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx.$

Hence
$\int_{-\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx.$

Note that
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=0$
since the fuction $\frac{\sin ^3 (\pi x)}{1+x^{10}}$ is odd.

Erdos
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