1 is a usubstitution . I don ' t change the bounds of integration , because I keep messing them up . So let u 1 z . Then du dz , and z 1 u . Integral z 1 z 12 Integral 1 uu 12 du substitute u for 1 z Integral 1 uu 12 du Integral u 32 u 12 du Integral u 32 u 12 2 u 525 2 u 323 power rule , and since this is going to be a definite integral , I don ' t care about the constant of integration 2 u 525 2 u 323 21 z 525 21 z 323 substitute 1 z back for u , and now you can evaluate going from z 0 to 1 Integration by parts Integral u dv uv Integral v du . u x du dx choose u so that it gets nicer when you take the derivative dv cos 3 x dx v sin 3 x 3 choose v so that it does not get messier when you integrate ituv Integral v du xsin 3 x 3 Integral sin 3 x 3 dx xsin 3 x 3 Integral sin 3 x 3 dx xsin 3 x 3 cos 3 x 9 xsin 3 x 3 cos 3 x 9 xsin 3 x 3 cos 3 x 9