What is the integral of (3x+6)cos(3x^2+12x) ?

The integral of (3x+6)cos(3x^2+12x) is 1/2 sin(3x^2+12x)+C

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integral of (3x+6)cos(3x^2+12x)

\int (3 x + 6) cos (3 x^{2} + 12 x) d x

\frac{1}{2} sin (3 x^{2} + 12 x) + C

Solution steps

\int (3 x + 6) cos (3 x^{2} + 12 x) d x

Apply u-substitution

: \int \frac{cos ( u )}{2} d u

= \int \frac{cos ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{2} sin (u)

Substitute back

u = 3 x^{2} + 12 x

= \frac{1}{2} sin (3 x^{2} + 12 x)

Add a constant to the solution

= \frac{1}{2} sin (3 x^{2} + 12 x) + C

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What is the integral of (3x+6)cos(3x^2+12x) ?
The integral of (3x+6)cos(3x^2+12x) is 1/2 sin(3x^2+12x)+C