Correct Answer - A
`sin nx-sin(n-2)x=2 cos (n-1)x sinx `
or `int(sin nx)/(sin x)dx=int2cos(n-1)dx+int(sin(n-2)x)/(sinx)dx`
`:. int_(0)^(pi//2) (sin 5x)/(sinx)dx=int_(0)^(pi//2) 2 cos4xdx+int_(0)^(pi//2) (sin3x)/(sinx) dx`
`=0+int_(0)^(pi//2)(sin 3x)/(sin x) dx`
`=int_(0)^(pi//2) dx`
`=(pi)/2`