Evaluate the following integral ∫_0^(π⁄4)cosx cos2x cos3x dx
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Mathematics
Since cosx cos2x cos3x=1/2 cosx [cos5x+cosx ]=1/2 [cosx cos5x+〖cos〗^2 x]
need an explanation for this answer? contact us directly to get an explanation for this answer=1/4 [cos6x+cos4x+1+cos2x ]
∴ ∫_0^(π⁄4)〖cosx cos2x cos3x dx〗=1/4 ∫_0^(π⁄4)(1+cos2x+cos4x cos6x )dx
=1/4 [x+1/2 sin2x+1/4 sin4x+1/6 sin6x ]_0^(π⁄4)
=1/4 [π/4+1/2 sin〖π/2〗+1/4 sinπ+1/6 〖sin〗^3 π/2]=1/4 [π/4+1/2+1/6]=(3π+4)/48