What is the integral of-6xsin(3x^2-1) ?

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

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integral of-6xsin(3x^2-1)

\int - 6 x sin (3 x^{2} - 1) d x

cos (3 x^{2} - 1) + C

Solution steps

\int - 6 x sin (3 x^{2} - 1) d x

Take the constant out:

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

= - 6 \cdot \int x sin (3 x^{2} - 1) d x

Apply u-substitution

: \int \frac{sin ( 3 u - 1 )}{2} d u

= - 6 \cdot \int \frac{sin ( 3 u - 1 )}{2} d u

Take the constant out:

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

= - 6 \cdot \frac{1}{2} \cdot \int sin (3 u - 1) d u

Apply u-substitution

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

= - 6 \cdot \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

= - 6 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int sin (v) d v

Use the common integral:

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

= - 6 \cdot \frac{1}{2} \cdot \frac{1}{3} (- cos (v))

Substitute back

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

Simplify

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

= cos (3 x^{2} - 1)

Add a constant to the solution

= cos (3 x^{2} - 1) + C

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