What is the integral of-(2x+1)sin(3x^2+3x) ?

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

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integral of-(2x+1)sin(3x^2+3x)

\int - (2 x + 1) sin (3 x^{2} + 3 x) d x

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

Solution steps

\int - (2 x + 1) sin (3 x^{2} + 3 x) d x

Take the constant out:

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

= - \int (2 x + 1) sin (3 x^{2} + 3 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^{2} + 3 x

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

Simplify

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

Add a constant to the solution

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

\int_{1}^{e} ln (x) d x

n = 1 \sum \infty \frac{1}{n ^{n}}

\int sec^{2} (2 x t a) n^{5} 2 x d x

\int \frac{x}{3 - x} d x

\frac{e ^{x}}{3 + e ^{x}}

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