Question

How do you evaluate the integral `int xsin(3x^2+1)`?

Answer 1

`intxsin(3x^2+1)dx=-1/6cos(3x^2+1)+C`

Explanation:

`I=intxsin(3x^2+1)dx`

Let `u=3x^2+1`, so `du=6xcolor(white).dx`.

`I=1/6intsin(3x^2+1)(6xcolor(white).dx)`

`I=1/6intsin(u)du`

`I=-1/6cos(u)+C`

`I=-1/6cos(3x^2+1)+C`