Evaluate the following integral I=∫_0^π〖〖x sin〗^7 x〗 dx
I=∫_0^π〖〖x sin〗^7 x〗 dx=∫_0^π〖〖(π-x) sin〗^7 (π-x) 〗 dx=∫_0^π〖〖(π-x) sin〗^7 x〗 dx ∴2I=∫_0^π〖〖π sin〗^7 x〗 dx=2π ∫_0^(π/2)〖〖 sin〗^7 x〗 dx=2π.6.4.2/7.5.3=32π/35
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I=∫_0^π〖〖x sin〗^7 x〗 dx=∫_0^π〖〖(π-x) sin〗^7 (π-x) 〗 dx=∫_0^π〖〖(π-x) sin〗^7 x〗 dx
need an explanation for this answer? contact us directly to get an explanation for this answer∴2I=∫_0^π〖〖π sin〗^7 x〗 dx=2π ∫_0^(π/2)〖〖 sin〗^7 x〗 dx=2π.6.4.2/7.5.3=32π/35