What is the integral of xsin(3x^2+10) ?

The integral of xsin(3x^2+10) is -1/6 cos(3x^2+10)+C

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integral of xsin(3x^2+10)

\int x sin (3 x^{2} + 10) d x

- \frac{1}{6} cos (3 x^{2} + 10) + C

Solution steps

\int x sin (3 x^{2} + 10) d x

Apply u-substitution

: \int \frac{sin ( 3 u + 10 )}{2} d u

= \int \frac{sin ( 3 u + 10 )}{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 (3 u + 10) d u

Apply u-substitution

: \int sin (v) \frac{1}{3} d v

= \frac{1}{2} \cdot \int sin (v) \frac{1}{3} 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}{3} \cdot \int sin (v) d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{3} (- cos (v))

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} (- cos (3 x^{2} + 10))

Simplify

\frac{1}{2} \cdot \frac{1}{3} (- cos (3 x^{2} + 10)) : - \frac{1}{6} cos (3 x^{2} + 10)

= - \frac{1}{6} cos (3 x^{2} + 10)

Add a constant to the solution

= - \frac{1}{6} cos (3 x^{2} + 10) + C

\frac{d}{d x} (ln (x y z))

\frac{\partial}{\partial x} (arctan (x + y))

\frac{d}{d x} (x (x - 6)^{2})

\int \frac{1}{x a ^{2} + x ^{2}} d x

n = 3 \sum \infty \frac{2}{n ^{2} - 1}

What is the integral of xsin(3x^2+10) ?
The integral of xsin(3x^2+10) is -1/6 cos(3x^2+10)+C