What is the integral of 1/2 xcos(-3x^2+8) ?

The integral of 1/2 xcos(-3x^2+8) is -1/12 sin(-3x^2+8)+C

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integral of 1/2 xcos(-3x^2+8)

\int \frac{1}{2} x cos (- 3 x^{2} + 8) d x

- \frac{1}{12} sin (- 3 x^{2} + 8) + C

Solution steps

\int \frac{1}{2} x cos (- 3 x^{2} + 8) d x

Take the constant out:

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

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

Apply u-substitution

: \int \frac{cos ( - 3 u + 8 )}{2} d u

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{3} sin (- 3 x^{2} + 8))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{3} sin (- 3 x^{2} + 8)) : - \frac{1}{12} sin (- 3 x^{2} + 8)

= - \frac{1}{12} sin (- 3 x^{2} + 8)

Add a constant to the solution

= - \frac{1}{12} sin (- 3 x^{2} + 8) + C

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What is the integral of 1/2 xcos(-3x^2+8) ?
The integral of 1/2 xcos(-3x^2+8) is -1/12 sin(-3x^2+8)+C