Q.

|120π3∫0πx2sinxcosxsin4x+cos4x dx| is equal to ________ .                 [2024]


Ans.

(15)

 Let I=∫0πx2sinxcosxsin4x+cos4x dx

=∫0π/2sinxcosxsin4x+cos4x(x2-(π-x)2) dx

=∫0π/2sinxcosx (2πx-π2)sin4x+cos4x dx

=2π∫0π/2xsinxcosxsin4x+cos4x dx-π2∫0π/2sinxcosxsin4x+cos4x dx

=2π·π4∫0π/2sinxcosxsin4x+cos4x -π2∫0π/2sinxcosxsin4x+cos4x dx

=-π22∫0π/2sinxcosxsin4x+cos4x =-π22∫0π/2sinxcosx1-2sin2x+cos2x dx

=-π22∫0π/2sin2x1+cos22x dx

Put cos2x=t  ∴ I=-π22∫1-1-dt2(1+t2)

=-π22∫01dt1+t2=-π22(tan-1t)01=-π22(π4)=-π38

∴  |120π3∫0πx2sinxcosxsin4x+cos4x |=120π3×π38=15