Question

pls integrate sin2x.cos3x

Answer 1

⌠sin2x cos3xdx
=⌠2sinx cosx(4sin​3x-3cosx)dx
=8⌠sinx cos4xdx  -  6⌠sinx cosxdx
let cosx=t
    or -sinxdx=dt
so, 8⌠(-t4)dt+6tdt
=6t2/2 - 8t5/5
=3cos2x-(8/5)cos5x+c    (Ans)