Intergration by substitution

solve using above method: sin*6/y / y^3
i am not sure how to do this

I think you were trying to write: sin(6/y) / y^3, if so

let u = 6 / y, then du = -6 / y^2 dy

then

sin(6/y) / y^3 dy becomes

-sin(u) du / u

This is the "Sine integral" Si(x) where x is the upper limit on your integration. It has no indefinite result.