What is the integral of xsin(6x^2+2) ?

The integral of xsin(6x^2+2) is -1/12 cos(6x^2+2)+C

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integral of xsin(6x^2+2)

\int x sin (6 x^{2} + 2) d x

- \frac{1}{12} cos (6 x^{2} + 2) + C

Solution steps

\int x sin (6 x^{2} + 2) d x

Apply u-substitution

: \int \frac{sin ( 6 u + 2 )}{2} d u

= \int \frac{sin ( 6 u + 2 )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sin (6 u + 2) d u

Apply u-substitution

: \int sin (v) \frac{1}{6} d v

= \frac{1}{2} \cdot \int sin (v) \frac{1}{6} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{6} \cdot \int sin (v) d v

Use the common integral:

\int sin (v) d v = - cos (v)

= \frac{1}{2} \cdot \frac{1}{6} (- cos (v))

Substitute back

= \frac{1}{2} \cdot \frac{1}{6} (- cos (6 x^{2} + 2))

Simplify

\frac{1}{2} \cdot \frac{1}{6} (- cos (6 x^{2} + 2)) : - \frac{1}{12} cos (6 x^{2} + 2)

= - \frac{1}{12} cos (6 x^{2} + 2)

Add a constant to the solution

= - \frac{1}{12} cos (6 x^{2} + 2) + C

\int \frac{1}{x ^{2} yz} d x

ln (4 - 3 x)

x y^{2} y^{^{'}} = x + 3

\int x^{3} ln (x^{3}) d x

y^{^{'}} = \frac{1}{sec ^{2} ( y )}

What is the integral of xsin(6x^2+2) ?
The integral of xsin(6x^2+2) is -1/12 cos(6x^2+2)+C