Q.

limxπ2(x3(π/2)3(sin(2t1/3)+cos(t1/3))dt(x-π2)2) is equal to            [2024]

1 11π210    
2 3π22    
3 5π29    
4 9π28  

Ans.

(4)

Let L=limxπ/2[x3(π/2)3(sin(2t1/3)+cos(t1/3))dt(x-π2)2] 00 form, so by L'Hospital rule

=limxπ2ddxx3(π/2)3(sin(2t1/3)+cos(t1/3))dt2(x-π2) sin(2×π2)·ddx(π2)3-sin(2x)·ddxx3

=limxπ/2+(cosπ2ddx(π2)3-cosx·ddxx3)2(x-π2)    (By Leibnitz rule of integration)

=limxπ/2-3x2sin2x-3x2cosx2(x-π2)  (00 form)

=limxπ/2-6xsin2x-6x2cos2x-6xcosx+3x2sinx2

=6π24+3π242=9π28