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

##### 1 Answer
Jan 20, 2017

$\int x \sin \left(3 {x}^{2} + 1\right) \mathrm{dx} = - \frac{1}{6} \cos \left(3 {x}^{2} + 1\right) + C$

#### Explanation:

$I = \int x \sin \left(3 {x}^{2} + 1\right) \mathrm{dx}$

Let $u = 3 {x}^{2} + 1$, so $\mathrm{du} = 6 x \textcolor{w h i t e}{.} \mathrm{dx}$.

$I = \frac{1}{6} \int \sin \left(3 {x}^{2} + 1\right) \left(6 x \textcolor{w h i t e}{.} \mathrm{dx}\right)$

$I = \frac{1}{6} \int \sin \left(u\right) \mathrm{du}$

$I = - \frac{1}{6} \cos \left(u\right) + C$

$I = - \frac{1}{6} \cos \left(3 {x}^{2} + 1\right) + C$