What is the integral of 12x^2sin(6-x^3) ?

The integral of 12x^2sin(6-x^3) is 4cos(6-x^3)+C

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integral of 12x^2sin(6-x^3)

\int 12 x^{2} sin (6 - x^{3}) d x

4 cos (6 - x^{3}) + C

Solution steps

\int 12 x^{2} sin (6 - x^{3}) d x

Take the constant out:

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

= 12 \cdot \int x^{2} sin (6 - x^{3}) d x

Apply u-substitution

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

= 12 \cdot \int - \frac{sin ( u )}{3} d u

Take the constant out:

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

= 12 (- \frac{1}{3} \cdot \int sin (u) d u)

Use the common integral:

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

= 12 (- \frac{1}{3} (- cos (u)))

Substitute back

u = 6 - x^{3}

= 12 (- \frac{1}{3} (- cos (6 - x^{3})))

Simplify

12 (- \frac{1}{3} (- cos (6 - x^{3}))) : 4 cos (6 - x^{3})

= 4 cos (6 - x^{3})

Add a constant to the solution

= 4 cos (6 - x^{3}) + C

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What is the integral of 12x^2sin(6-x^3) ?
The integral of 12x^2sin(6-x^3) is 4cos(6-x^3)+C