What is the integral of-8xsin(4x^2-3) ?

The integral of-8xsin(4x^2-3) is cos(4x^2-3)+C

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integral of-8xsin(4x^2-3)

\int - 8 x sin (4 x^{2} - 3) d x

cos (4 x^{2} - 3) + C

Solution steps

\int - 8 x sin (4 x^{2} - 3) d x

Take the constant out:

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

= - 8 \cdot \int x sin (4 x^{2} - 3) d x

Apply u-substitution

: \int \frac{sin ( 4 u - 3 )}{2} d u

= - 8 \cdot \int \frac{sin ( 4 u - 3 )}{2} d u

Take the constant out:

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

= - 8 \cdot \frac{1}{2} \cdot \int sin (4 u - 3) d u

Apply u-substitution

: \int sin (v) \frac{1}{4} d v

= - 8 \cdot \frac{1}{2} \cdot \int sin (v) \frac{1}{4} d v

Take the constant out:

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

= - 8 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int sin (v) d v

Use the common integral:

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

= - 8 \cdot \frac{1}{2} \cdot \frac{1}{4} (- cos (v))

Substitute back

= - 8 \cdot \frac{1}{2} \cdot \frac{1}{4} (- cos (4 x^{2} - 3))

Simplify

- 8 \cdot \frac{1}{2} \cdot \frac{1}{4} (- cos (4 x^{2} - 3)) : cos (4 x^{2} - 3)

= cos (4 x^{2} - 3)

Add a constant to the solution

= cos (4 x^{2} - 3) + C

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What is the integral of-8xsin(4x^2-3) ?
The integral of-8xsin(4x^2-3) is cos(4x^2-3)+C