The value of limx→0 ∫0x2 cost2dtxsinx, is
3/2
1
-1
none of these
We have,
limx→0 ∫0x2 cost2dtxsinx=limx→0 2xcosx4xcosx+sinx [Using L'Hospital's Rule] =limx→0 2cosx4−8x4sinx42cosx−xsinx=2−02−0=1