Integral equality  

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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 Dr Ali
    +2

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1 Answer

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 Erdos
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