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

The integral of 3x^2sin(4x^3-2) is -1/4 cos(4x^3-2)+C

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integral of 3x^2sin(4x^3-2)

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

- \frac{1}{4} cos (4 x^{3} - 2) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \frac{1}{3} \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

= 3 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \int sin (v) d v

Use the common integral:

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

= 3 \cdot \frac{1}{3} \cdot \frac{1}{4} (- cos (v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{4} cos (4 x^{3} - 2) + C

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What is the integral of 3x^2sin(4x^3-2) ?
The integral of 3x^2sin(4x^3-2) is -1/4 cos(4x^3-2)+C