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

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

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Answer 1 · 2017-01-20T03:27:38

$\int x sin (3 x^{2} + 1) d x = - \frac{1}{6} cos (3 x^{2} + 1) + C$ $intxsin(3x^2+1)dx=-1/6cos(3x^2+1)+C$

Explanation:

$I = \int x sin (3 x^{2} + 1) d x$ $I=intxsin(3x^2+1)dx$

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

$I = \frac{1}{6} \int sin (3 x^{2} + 1) (6 x . d x)$ $I=1/6intsin(3x^2+1)(6xcolor(white).dx)$

$I = \frac{1}{6} \int sin (u) d u$ $I=1/6intsin(u)du$

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

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

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

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Explanation:

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