Prove that [tex] \int_0^a f(x)dx = \int_0^{\frac{a}{2}}[f(x) + f(a-x)]dx \\[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Hence prove that [tex] \int_0^{\pi} x sin^6x dx = \pi \int_0^{\frac{\pi}{2}} sin^6x dx \\ [/tex] and evaluate this integral using the following reduction formula [tex] I_{m,n}= \frac{n-1}{m+n}I_{m,n-2} \\ [/tex]

My effort:

[tex] \int_0^a f(x)dx = \int_0^{\frac{a}{2}}f(x)dx + \int_{\frac{a}{2}} f(a-x)dx \\ [/tex]. Now

[tex] \int_{\frac{a}{2}}^a f(x)dx = \int_{\frac{a}{2}}^a f(a-x)dx \\ [/tex] ... by a theorem. Therefore [tex] \int_0^a f(x)dx = \int_0^{\frac{a}{2}} f(x)dx + \int_{\frac{a}{2}}^a f(a-x)dx \\ [/tex]. I need help to change the integral's limits to 0 and a/2 on the second integral on the right, thanks very much.

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# Homework Help: Simple proof needed

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