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

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

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integral of x^2cos(x^3)3sin(x^3)

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

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

Solution steps

\int x^{2} cos (x^{3}) \cdot 3 sin (x^{3}) 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} cos (x^{3}) sin (x^{3}) d x

Rewrite using trig identities

= 3 \cdot \int \frac{1}{2} x^{2} sin (2 x^{3}) d x

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int x^{2} sin (2 x^{3}) d x

Apply u-substitution

: \int \frac{sin ( u )}{6} d u

= 3 \cdot \frac{1}{2} \cdot \int \frac{sin ( u )}{6} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{6} (- cos (u))

Substitute back

u = 2 x^{3}

= 3 \cdot \frac{1}{2} \cdot \frac{1}{6} (- cos (2 x^{3}))

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{6} (- cos (2 x^{3})) : - \frac{1}{4} cos (2 x^{3})

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

Add a constant to the solution

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

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