Đặt u=x−π3⇒du=dx,u∈[−π3,0]π3∫0sinxcos3(x−π3)dx=0∫−π3sin(u+π3)cos3udu=0∫−π312sinu+√32cosucos3udu
=120∫−π3sinu.ducos3u+√320∫−π3ducos2u
=120∫−π3−d(cosu)cos3u+√320∫−π3tanu
=140∫−π31cos2u+√32(tan0−tan−π3)
=14(1cos20−1cos2−π3)+√32√3
=14(1−4)+32
=−34+32
=34