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

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Answer 1 · 2017-01-20T03:27:38

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

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

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

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

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

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

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

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