Integral of (x^3*sin(x^4)) dx?

Question

How do I solve it?

Answer 1

Let y = ∫ x^3*sin(x^4) dx

Let us assume that u = x^4

                               du = 4 x^3 dx

                               x^3 dx = (1/4) du

y = (1/4) ∫ sin(u) du

y = - cos(u) /4 + C                     [ Since ∫ sin(u) du = - cos(u) + c ]

Therefore ∫ x^3*sin(x^4) dx = - cos(x^4) /4 + C