What is the integral of (3x^2+3)sin(3x^3+9x) ?

The integral of (3x^2+3)sin(3x^3+9x) is -1/3 cos(3x^3+9x)+C

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integral of (3x^2+3)sin(3x^3+9x)

\int (3 x^{2} + 3) sin (3 x^{3} + 9 x) d x

- \frac{1}{3} cos (3 x^{3} + 9 x) + C

Solution steps

\int (3 x^{2} + 3) sin (3 x^{3} + 9 x) d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

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

Substitute back

u = 3 x^{3} + 9 x

= \frac{1}{3} (- cos (3 x^{3} + 9 x))

Simplify

= - \frac{1}{3} cos (3 x^{3} + 9 x)

Add a constant to the solution

= - \frac{1}{3} cos (3 x^{3} + 9 x) + C

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What is the integral of (3x^2+3)sin(3x^3+9x) ?
The integral of (3x^2+3)sin(3x^3+9x) is -1/3 cos(3x^3+9x)+C