Verified answer

∫ x²cos(x^3) dx

Let u = x^3

=> du = 3x² dx

<=> du/3 = x²dx

=1/3 ∫ cosu du

  sinu

= ▬▬▬ + C

   3

 sinx^3

=▬▬▬▬ + C

   3

If the integrand is of the type f(x)g(y), then you could in basic terms chop up the integrals into their respective variables, so ?[0,?/4] ?[0,?/2] sin(x)cos(2y) dx dy = (?[0,?/2] sin(x) dx)(?[0,?/4] cos(2y) dy) the two a form of contain in basic terms your well-known run of the mill single variable essential.