Integral of sin(πr)rdr: Step-by-Step Solution using Integration by Parts

To evaluate the integral \u00e2\u0088\u008fsin(\u00cf\u008fr)rdr, we can use integration by parts. \u000a\u000aLet's choose u = r and dv = sin(\u00cf\u008fr)dr. \u000a\u000aThen, du = dr and v = -1/\u00cfcos(\u00cf\u008fr). \u000a\u000aUsing the integration by parts formula, we have:\u000a\u000a\u00e2\u0088\u008fsin(\u00cf\u008fr)rdr = uv - \u00e2\u0088\u008fvdu\u000a\u000a\u000a = -1/\u00cf rcos(\u00cf\u008fr) - \u00e2\u0088\u008f(-1/\u00cfcos(\u00cf\u008fr))dr\u000a\u000a\u000a = -1/\u00cf rcos(\u00cf\u008fr) + 1/\u00cf\u00e2\u0088\u008fcos(\u00cf\u008fr)dr\u000a\u000a\u000a = -1/\u00cf rcos(\u00cf\u008fr) + 1/\u00cf(1/\u00cfsin(\u00cf\u008fr))\u000a\u000a\u000a = -1/\u00cf rcos(\u00cf\u008fr) + 1/\u00cf^2sin(\u00cf\u008fr) + C\u000a\u000aSo, the integral of sin(\u00cf\u008fr)rdr is equal to -1/\u00cf rcos(\u00cf\u008fr) + 1/\u00cf^2sin(\u00cf\u008fr) + C, where C is the constant of integration.