Question 861010
{{{u=3x^2+4x}}}
{{{du=(6x+4)dx}}}
So then,
{{{int(cos(3x^2+4x)*(6x+4),dx)=int(cos(u),du)}}}
{{{int(cos(3x^2+4x)*(6x+4),dx)=sin(u)+C}}}
{{{int(cos(3x^2+4x)*(6x+4),dx)=sin(3x^2+4x)+C}}}