Evaluate the following integrals as limit of sum: `int_0^(pi//2)sinx\ dx`

Explanation

`int_0^(pi//2)sinx\ dx`
we know,
`int_a^b f(x)dx =(b-a) lim_(n->oo)1/n[f(a)+f(a+h)+f(a+2h)+....+f(a+(n-1)h)`
where `a=0,b=pi/2`
`h=(b-a)/h`
`=>h= pi/2n`
`int_0^(pi/2) sin x dx =lim_(h->0) h[f(0)+f(h)+f(2h)+....+f(n-1)h]`
`=lim_(h->0) h[sin 0+sin h +sin 2h +.....+sin(n-1)h]`
`=lim_(h->0) h[(sin ((nh-h)/2) sin (nh)/2 )/ sin(h/2) ]`
`=lim_(h->0) h[(sin (pi/2-h)/2 cos (pi/4) )/ sin (h/2) ]`
`=lim_(h->0) h[(sin(pi/4-h/2)sin pi/4)/cos (h/2)]`
`=lim_(h->0) h/sin(h/2)[1/sqrt2 xx 1/sqrt2 ]`
`=1/2 lim_(h->0) (h/2 xx 2)/sin(h/2)`
`=1/2 xx 2`
`=1`