1I am getting the ans. 0. 1st evaluate the indefinite integral, then put the limits & the reqd condition.
∫sin(ax)sin(bx)dx = 12∫cos(a-b)x - cos(a+b)x
=1/2[sin(a-b)x(a-b) - sin(a+b)x(a+b)]
putting limits 0 to 1 we have,
12[sin(a-b)(a-b) - sin(a+b)(a+b)]
now, 2(a-b)= tana - tanb & 2(a+b)=tana + tanb
so, we have,
sin(a-b)tana - tanb - sin(a+b)tana + tanb
=cosa*cosb- cosa*cosb=0
