Question

Integral of x^3sin(x^4)dx​

Answer 1

Answer:

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 ]