What is the integral of x^2cos(4x^3-1) ?

The integral of x^2cos(4x^3-1) is 1/12 sin(4x^3-1)+C

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integral of x^2cos(4x^3-1)

\int x^{2} cos (4 x^{3} - 1) d x

\frac{1}{12} sin (4 x^{3} - 1) + C

Solution steps

\int x^{2} cos (4 x^{3} - 1) d x

Apply u-substitution

: \int \frac{u cos ( 4 u ^{\frac{3}{2}} - 1 )}{2 x} d u

= \int \frac{cos ( 4 u - 1 )}{3} d u

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{1}{4} sin (4 x^{3} - 1) : \frac{1}{12} sin (4 x^{3} - 1)

= \frac{1}{12} sin (4 x^{3} - 1)

Add a constant to the solution

= \frac{1}{12} sin (4 x^{3} - 1) + C

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What is the integral of x^2cos(4x^3-1) ?
The integral of x^2cos(4x^3-1) is 1/12 sin(4x^3-1)+C